Explanation:
Okay, so I need to find the area of the quadrilateral formed by the four points of tangency between the circle x² + y² = 2 and the parabola y² = 8x. The problem mentions there are two common tangents, forming four points of tangency. First, I need to figure out what these common tangents are, find their points of tangency on both the circle and the parabola, and then determine the quadrilateral they form. Finally, I have to calculate its area.

Let me start by recalling that a common tangent to two curves is a line that touches both curves at exactly one point each. Since the circle and the parabola are given, I need to find equations for these common tangents.

First, let me write down the equations again:

Circle: x² + y² = 2.

Parabola: y² = 8x.

I remember that the general equation of a tangent to a parabola can be written in terms of slope. For the parabola y² = 4ax, the tangent is y = mx + a/m. Comparing this to our parabola y² = 8x, which is equivalent to 4a = 8, so a = 2. Therefore, the tangent to the parabola y² = 8x can be written as y = mx + 2/m. Wait, let me verify that.

Alternatively, for a general parabola y² = 4ax, the equation of tangent with slope m is y = mx + a/m. So yes, in this case, since 4a = 8, a = 2, so the tangent line is y = mx + 2/m. So that's the equation of the tangent line to the parabola with slope m.

Now, this line should also be tangent to the circle x² + y² = 2. The condition for a line y = mx + c to be tangent to the circle x² + y² = r² is that the distance from the center of the circle to the line equals the radius.

The circle here is centered at (0, 0) with radius √2. So the distance from (0, 0) to the line y = mx + c is |0 - m*0 - c| / sqrt(m² + 1) ) = |c| / sqrt(m² + 1). For this to be equal to the radius √2, we have |c| / sqrt(m² + 1) = √2. Therefore, |c| = √2 * sqrt(m² + 1). Therefore, c = ±√2 * sqrt(m² + 1).

But since for the parabola's tangent line, we have c = 2/m. Therefore, setting these equal:

For the line to be tangent to both the parabola and the circle, we must have:

2/m = ±√2 * sqrt(m² + 1)

So this equation can be used to solve for m.

Let me write that equation:

2/m = ±√2 * sqrt(m² + 1)

Let me square both sides to eliminate the square root and the absolute value. But before that, let me note that m cannot be zero because the denominator would be zero. Also, since the parabola y² = 8x opens to the right, the tangent lines might have positive or negative slopes. Let me proceed step by step.

So squaring both sides:

(2/m)² = (√2 * sqrt(m² + 1))²

Which simplifies to:

4/m² = 2*(m² + 1)

Multiply both sides by m²:

4 = 2*(m² + 1)*m²

Divide both sides by 2:

2 = (m² + 1)*m²

So:

m⁴ + m² - 2 = 0

Let me set z = m², then the equation becomes:

z² + z - 2 = 0

Solving this quadratic equation:

z = [-1 ± sqrt(1 + 8)] / 2 = [-1 ± 3]/2

Therefore, z = (-1 + 3)/2 = 2/2 = 1 or z = (-1 - 3)/2 = -4/2 = -2

But since z = m², it can't be negative. Therefore, z = 1, so m² = 1 => m = ±1

Therefore, the slopes are m = 1 and m = -1.

But wait, earlier we had 2/m = ±√2 * sqrt(m² + 1). Let me check if these slopes satisfy the original equation.

Let me check m = 1:

Left side: 2/m = 2/1 = 2

Right side: ±√2 * sqrt(1 + 1) = ±√2 * √2 = ±2

Therefore, 2 = ±2. So the positive sign. So m = 1 gives c = 2/m = 2. Therefore, equation y = x + 2.

Similarly, for m = -1:

Left side: 2/m = 2/(-1) = -2

Right side: ±√2 * sqrt(1 + 1) = ±2. So we need -2 = ±2, which implies that we take the negative sign. Therefore, c = -2. Therefore, the equation is y = -x - 2.

Wait, but let me check if these lines are actually tangent to the parabola. For example, for m = 1, the line is y = x + 2. Let me check if this intersects the parabola y² = 8x.

Substitute y = x + 2 into y² = 8x:

(x + 2)² = 8x => x² + 4x + 4 = 8x => x² - 4x + 4 = 0 => (x - 2)² = 0. Therefore, x = 2, and y = 2 + 2 = 4. So yes, it's tangent at (2, 4). Similarly, for m = -1, the line is y = -x - 2. Substitute into y² = 8x:

(-x - 2)^2 = 8x => x² + 4x + 4 = 8x => x² - 4x + 4 = 0 => (x - 2)^2 = 0. Therefore, x = 2, y = -2 -2 = -4. Therefore, tangent at (2, -4).

Wait, but the problem mentions that there are two common tangents forming four points of tangency. But according to this, each tangent touches the parabola at one point and the circle at one point. So each tangent has two points of tangency? Wait, but the problem says "forming four points of tangency". So maybe there are two tangents, each tangent to both the circle and the parabola, so each tangent touches both curves once, so total four points of tangency? That seems possible. However, the Asymptote code in the problem shows four points A, B, C, D, with A = (-1,1), B = (2,4), C = (-1,-1), D = (2,-4). Wait, but according to my calculation, the tangents touch the parabola at (2,4) and (2,-4). Where are the other points?

Wait, maybe I made a mistake here. Maybe there are more common tangents?

Wait, let me think again. The equations of the common tangents are y = x + 2 and y = -x - 2. However, maybe there are also other common tangents?

Wait, but according to the algebra, solving for m led us to only two possible slopes, m = ±1. However, the Asymptote code shows that there are four points of tangency, with coordinates like (-1,1), (-1,-1), (2,4), (2,-4). So maybe there are two common tangents, each tangent to the circle and parabola, but each tangent is touching the circle at two points? No, a line can only be tangent to a circle at one point. So perhaps there are two common tangents, but each tangent line touches the circle and the parabola once, hence four points of tangency: two on the circle and two on the parabola? Wait, but in that case, the Asymptote code shows that the points of tangency on the parabola are (2,4) and (2,-4), as we found, and the points of tangency on the circle would be... Let me find them.

So for the line y = x + 2, which touches the parabola at (2, 4). Now, let's find the point where this line is tangent to the circle. For that, we need to find the point on the circle where the line is tangent.

Alternatively, since we know that the tangent to the circle is y = x + 2. The tangent line to a circle at a point (x1, y1) is xx1 + yy1 = r². Here, the circle is x² + y² = 2, so the tangent at (x1, y1) is xx1 + yy1 = 2. Let me compare that with the line y = x + 2.

So, let me write the equation of the tangent line to the circle as y = x + 2. Let me rearrange this to standard form: x - y + 2 = 0. Wait, the standard form for the tangent line to the circle is xx1 + yy1 = 2. So equating coefficients?

Alternatively, suppose the tangent line is y = x + 2. Then, since it's tangent to the circle x² + y² = 2, the point of tangency (x1, y1) lies on the line and satisfies the condition that the vector from the center to (x1, y1) is perpendicular to the tangent line.

The slope of the tangent line is 1, so the slope of the radius is -1 (negative reciprocal). Therefore, the line from (0,0) to (x1, y1) has slope y1/x1 = -1, so y1 = -x1. Also, since (x1, y1) lies on the line y = x + 2, substituting y1 = -x1 into that gives -x1 = x1 + 2 → -2x1 = 2 → x1 = -1. Therefore, x1 = -1, y1 = -(-1) = 1. Therefore, the point of tangency on the circle is (-1, 1). Similarly, for the other tangent line y = -x - 2, let's find the point of tangency.

Same process. The tangent line has slope -1, so the radius has slope 1. Therefore, y1/x1 = 1 ⇒ y1 = x1. Since the point lies on the line y = -x - 2, substituting y1 = x1 gives x1 = -x1 - 2 ⇒ 2x1 = -2 ⇒ x1 = -1, so y1 = -1. Therefore, the point of tangency on the circle is (-1, -1). Therefore, the four points of tangency are (-1, 1) on the circle, (2, 4) on the parabola, (-1, -1) on the circle, and (2, -4) on the parabola. Then, these four points form a quadrilateral. The Asymptote code also labels these points as A(-1,1), B(2,4), C(-1,-1), D(2,-4). Then, connecting them as A-B-C-D-A? Wait, how is the quadrilateral formed? The Asymptote code draws lines connecting A to C and B to D. So maybe the quadrilateral is A-B-D-C or A-B-D-C? Wait, looking at the Asymptote code:

draw(interp(A,B,-0.2)--interp(A,B,1.2)); which is the tangent line AB?

Wait, the Asymptote code is a bit unclear, but the key is that the four points of tangency are A(-1,1), B(2,4), C(-1,-1), D(2,-4). So connecting these four points, what quadrilateral do they form? Let me plot them mentally.

Point A(-1,1) is on the left side of the circle, upper half. Point B(2,4) is on the parabola, far right upper. Point C(-1,-1) is on the left side, lower half. Point D(2,-4) is on the parabola, far right lower. So connecting A-B-D-C-A? Let me think. If we connect A to B to D to C to A, that would form a quadrilateral. Alternatively, maybe A-B-C-D? Wait, but C is (-1,-1), so connecting A(-1,1) to B(2,4) to C(-1,-1) would cross over. Maybe it's a kite-shaped figure or some quadrilateral.

Alternatively, since the tangent lines are AB and CD? Wait, the tangent lines are AB: y = x + 2, which touches the parabola at B(2,4) and the circle at A(-1,1). Similarly, the other tangent line is y = -x - 2, which touches the parabola at D(2,-4) and the circle at C(-1,-1). Therefore, the two tangent lines are AB and CD. But how do these points form a quadrilateral? Maybe the quadrilateral is formed by connecting A to B to D to C to A? Let me check coordinates:

A(-1,1), B(2,4), D(2,-4), C(-1,-1). Connecting A to B to D to C to A. Let me visualize:

- From (-1,1) to (2,4): upper right
- From (2,4) to (2,-4): vertical line down the right side
- From (2,-4) to (-1,-1): diagonal left up
- From (-1,-1) back to (-1,1): vertical line up the left side.

Wait, that seems to form a trapezoid? Maybe not. Alternatively, maybe it's a quadrilateral with two sides on the tangent lines and the other sides as connecting the points? Wait, but the problem says "the quadrilateral formed by the four points of tangency", so it's the quadrilateral with vertices at those four points. So regardless of how they're connected, we can compute the area.

But first, let me make sure the four points are correct. So we have:

For the tangent line y = x + 2:

- Point of tangency with circle: (-1, 1)
- Point of tangency with parabola: (2, 4)

For the tangent line y = -x - 2:

- Point of tangency with circle: (-1, -1)
- Point of tangency with parabola: (2, -4)

So these four points: (-1,1), (2,4), (-1,-1), (2,-4). So plotting these, they form a quadrilateral. To find the area, perhaps we can use coordinates to compute the area.

Let me note the coordinates again:

A(-1,1), B(2,4), C(-1,-1), D(2,-4). Wait, but in the Asymptote code, they have A(-1,1), B(2,4), C(-1,-1), D(2,-4). So connecting A, B, D, C? Let me check if that's a quadrilateral. Alternatively, maybe A, B, C, D? But connecting A(-1,1) to B(2,4) to C(-1,-1) to D(2,-4) would cross over. Alternatively, maybe the quadrilateral is A, B, D, C connected as a trapezoid? Let me see.

Alternatively, since there are two tangent lines, each tangent line has two points of tangency. So maybe the quadrilateral is formed by connecting the two points on the circle with the two points on the parabola? But how?

Alternatively, maybe the quadrilateral is a rectangle or some symmetric figure. Let me check the coordinates.

Looking at points A(-1,1), B(2,4), C(-1,-1), D(2,-4). If I think about these points, they are symmetric with respect to both the x-axis and y-axis? Let me check:

Point A(-1,1) and point C(-1,-1) are symmetric over the x-axis.

Point B(2,4) and D(2,-4) are symmetric over the x-axis.

Also, if we consider the y-axis, point A(-1,1) and the point (1,1) would be symmetric, but there is no such point here. Similarly for others.

But perhaps the quadrilateral is symmetric over the x-axis. If that's the case, then maybe I can compute the area of the upper half and double it.

Alternatively, use coordinates to compute the area using the shoelace formula.

Yes, the shoelace formula is a good method for finding the area of a polygon given its coordinates. However, to use that, I need to ensure the points are ordered either clockwise or counterclockwise.

Let me order the points properly. Let's take A(-1,1), B(2,4), D(2,-4), C(-1,-1), and back to A(-1,1). Let me verify if this order is correct.

Plotting the points:

Starting at (-1,1) (A), going to (2,4) (B), then to (2,-4) (D), then to (-1,-1) (C), then back to A. This should create a quadrilateral. Let me check if this is a trapezoid or some other shape. Alternatively, perhaps it's a parallelogram? Let me check the slopes of the sides.

From A(-1,1) to B(2,4): slope is (4 - 1)/(2 - (-1)) = 3/3 = 1.

From B(2,4) to D(2,-4): slope is (-4 - 4)/(2 - 2) = (-8)/0, undefined (vertical line).

From D(2,-4) to C(-1,-1): slope is (-1 - (-4))/(-1 - 2) = (3)/(-3) = -1.

From C(-1,-1) to A(-1,1): slope is (1 - (-1))/(-1 - (-1)) = 2/0, undefined (vertical line).

So the sides are: two sides with slope 1 and -1, and two vertical sides. Therefore, the quadrilateral is a trapezoid with two vertical sides and two non-parallel sides. Wait, but in trapezoid, only one pair of sides is parallel. However, here, the sides AB and CD have slopes 1 and -1, which are negative reciprocals, so they are not parallel. The vertical sides are parallel to each other (both vertical). So yes, this is a trapezoid with the two vertical sides as the parallel sides.

Therefore, the area can be calculated as the average of the lengths of the two parallel sides multiplied by the distance between them.

The two vertical sides are from A(-1,1) to C(-1,-1) and from B(2,4) to D(2,-4). The lengths of these sides:

Left side: from (-1,1) to (-1,-1): length is |1 - (-1)| = 2.

Right side: from (2,4) to (2,-4): length is |4 - (-4)| = 8.

The distance between these two vertical sides is the horizontal distance between x = -1 and x = 2, which is 2 - (-1) = 3.

Therefore, the area would be average of 2 and 8 multiplied by 3: ((2 + 8)/2) * 3 = (10/2) * 3 = 5 * 3 = 15.

Wait, but let me verify this with shoelace formula to make sure.

Alternatively, using shoelace formula for coordinates ordered correctly.

Order of points: A(-1,1), B(2,4), D(2,-4), C(-1,-1), back to A(-1,1).

Shoelace formula:

Area = 1/2 |sum_{i=1 to n} (x_i y_{i+1} - x_{i+1} y_i)|

Let me compute each term:

First, list the coordinates in order:

1. (-1, 1)
2. (2, 4)
3. (2, -4)
4. (-1, -1)
5. (-1, 1) [closing the polygon]

Compute x_i y_{i+1} - x_{i+1} y_i for each i from 1 to 4.

Term 1 (i=1): x1 y2 - x2 y1 = (-1)(4) - (2)(1) = -4 - 2 = -6

Term 2 (i=2): x2 y3 - x3 y2 = (2)(-4) - (2)(4) = -8 - 8 = -16

Term 3 (i=3): x3 y4 - x4 y3 = (2)(-1) - (-1)(-4) = -2 - 4 = -6

Term 4 (i=4): x4 y5 - x5 y4 = (-1)(1) - (-1)(-1) = -1 - 1 = -2

Sum of terms: -6 + (-16) + (-6) + (-2) = -30

Take absolute value and multiply by 1/2: 1/2 * | -30 | = 15. So area is 15. Which matches the previous calculation.

Therefore, the area of the quadrilateral is 15. Wait, but let me check again if the order of points is correct. Because sometimes depending on the order, the shoelace formula can give a wrong area if the points are not ordered correctly.

Let me try another order. Suppose we take A(-1,1), B(2,4), C(-1,-1), D(2,-4), back to A. Wait, would that give a different area?

Let me compute:

Coordinates:

1. (-1,1)
2. (2,4)
3. (-1,-1)
4. (2,-4)
5. (-1,1)

Compute terms:

Term 1: (-1)(4) - (2)(1) = -4 -2 = -6

Term 2: (2)(-1) - (-1)(4) = -2 + 4 = 2

Term 3: (-1)(-4) - (2)(-1) = 4 + 2 = 6

Term 4: (2)(1) - (-1)(-4) = 2 - 4 = -2

Sum: -6 + 2 + 6 - 2 = 0. That can't be right. So that order is not correct. Therefore, the correct order is A, B, D, C.

Alternatively, another order: A(-1,1), C(-1,-1), D(2,-4), B(2,4), back to A. Let me try.

Coordinates:

1. (-1,1)
2. (-1,-1)
3. (2,-4)
4. (2,4)
5. (-1,1)

Compute terms:

Term1: (-1)(-1) - (-1)(1) = 1 +1 = 2

Term2: (-1)(-4) - (2)(-1) = 4 +2 = 6

Term3: (2)(4) - (2)(-4) = 8 + 8 = 16

Term4: (2)(1) - (-1)(4) = 2 +4 = 6

Total sum: 2 +6 +16 +6 = 30. Area is 1/2*|30| = 15. Same result. So regardless of the order as long as it's correct, the area is 15. Therefore, seems like the area is 15.

But let me think again if this makes sense. The quadrilateral has two vertical sides of length 2 and 8, separated by 3 units horizontally. The area of trapezoid is average of the two bases times height. The bases are 2 and 8, height is 3. So yes, (2 + 8)/2 * 3 = 15. Therefore, seems correct.

But let me check with another approach. For example, coordinates of the points:

A(-1,1), B(2,4), C(-1,-1), D(2,-4). Alternatively, we can think of this quadrilateral as two triangles.

For instance, divide the quadrilateral into two triangles: ABC and ACD? Wait, but need to make sure the triangles are non-overlapping.

Alternatively, split into triangle ABD and triangle ACD? Maybe not. Let me think.

Alternatively, since the quadrilateral is symmetric over the x-axis, we can compute the area of the upper half and double it.

The upper half would be the area between points A(-1,1), B(2,4), and the reflection of D over x-axis which is (2,4). Wait, no, maybe not.

Wait, the quadrilateral is symmetric over the x-axis, since if you reflect over x-axis, A(-1,1) goes to C(-1,-1), B(2,4) goes to D(2,-4). So the upper half would be the triangle formed by A, B, and the midpoint?

Alternatively, since the quadrilateral is symmetric, we can compute the area of the upper half (the part above the x-axis) and double it.

The upper half would be the area bounded by A(-1,1), B(2,4), and the x-axis? Wait, not exactly. Wait, the quadrilateral includes points above and below the x-axis. But due to symmetry, maybe the upper half is the area between A, B, and the reflection of C?

Wait, maybe not. Let me instead compute the area of the trapezoid as a combination of two trapezoids or something else. However, since we already used two methods (trapezoid formula and shoelace formula) and both gave 15, which seems consistent, maybe that's the answer.

But just to be thorough, let me check once more using vectors or coordinate geometry.

Alternatively, consider coordinates:

The quadrilateral has vertices at (-1,1), (2,4), (2,-4), (-1,-1). Let me plot them mentally. The figure is a trapezoid with the two vertical sides at x=-1 and x=2. The left vertical side goes from (-1,1) to (-1,-1), which is a length of 2 units. The right vertical side goes from (2,4) to (2,-4), which is a length of 8 units. The horizontal distance between x=-1 and x=2 is 3 units. Then, as per trapezoid area formula, average of the two bases (2 and 8) multiplied by the height (distance between the vertical sides, which is 3). Therefore, area = ((2 + 8)/2)*3 = 5*3 = 15. Yep, same as before.

Alternatively, think of the figure as a rectangle with a certain width and height, but modified.

But since multiple methods give 15, and the coordinates are given in the Asymptote code, which might be a hint as well. The Asymptote code has A(-1,1), B(2,4), C(-1,-1), D(2,-4). So the area is indeed 15.

But let me check again if my calculation of the points of tangency is correct.

For the tangent line y = x + 2:

- On the circle x² + y² = 2. We found the point (-1,1). Let me verify that (-1)^2 + (1)^2 = 1 + 1 = 2, which is correct. So that point is on the circle.

- On the parabola y² = 8x. At point (2,4): 4² = 16, 8*2 = 16. Correct.

Similarly for the line y = -x - 2:

- On the circle: (-1, -1). (-1)^2 + (-1)^2 = 1 + 1 = 2. Correct.

- On the parabola: (2, -4). (-4)^2 = 16, 8*2 = 16. Correct.

Therefore, the points are correct. So the quadrilateral formed by these four points has area 15.

Wait, but the problem says "the quadrilateral formed by the four points of tangency." But according to the Asymptote code, they also drew lines connecting A to C and B to D, which are the vertical lines. That suggests that the quadrilateral is a rectangle? Wait, but according to the coordinates, it's not a rectangle. Because the sides AB and CD have slopes 1 and -1, which are not perpendicular. Wait, let me check angles.

Alternatively, maybe my assumption about the quadrilateral is wrong? Wait, the Asymptote code draws:

draw(interp(A,B,-0.2)--interp(A,B,1.2)); which is the tangent line AB.

Similarly for CD.

Then draw(A--C); draw(B--D); which connects A to C and B to D.

So the figure is formed by the two tangent lines and the two vertical lines connecting the circle points to the parabola points. But the four points are A, B, C, D, forming the quadrilateral. So the sides of the quadrilateral are AB, BD, DC, and CA? Wait, no. If you connect A to B to D to C to A, then the sides are AB, BD, DC, and CA. However, BD is from B(2,4) to D(2,-4), which is vertical line. DC is from D(2,-4) to C(-1,-1). And CA is from C(-1,-1) to A(-1,1). And AB is from A to B.

Wait, but in this case, the quadrilateral is a trapezoid with two vertical sides and two non-parallel sides, which matches our previous calculation.

But just to confirm, if I think of the coordinates, the quadrilateral is symmetric about the x-axis. Also, the area is 15. Let me check if the answer is 15. However, the problem says "the area of the quadrilateral formed by the four points of tangency." So 15 seems correct, but let me check with another method.

Alternatively, consider coordinates matrix:

For shoelace formula, we can write the coordinates in order:

(-1, 1), (2, 4), (2, -4), (-1, -1)

Applying shoelace formula step by step:

List the coordinates:

x: -1, 2, 2, -1, -1

y: 1, 4, -4, -1, 1

Compute sum of x_i y_{i+1}:

(-1)(4) + (2)(-4) + (2)(-1) + (-1)(1) =

(-4) + (-8) + (-2) + (-1) = -15

Compute sum of y_i x_{i+1}:

(1)(2) + (4)(2) + (-4)(-1) + (-1)(-1) =

2 + 8 + 4 + 1 = 15

Then subtract the two sums: |-15 - 15| = |-30| = 30. Then area is 1/2 * 30 = 15. Yep, same result. Therefore, confident that the area is 15.

However, just to be thorough, let me check if there are other common tangents which I might have missed. Earlier, when solving for the slopes, I obtained m = ±1. But is that all?

Let me revisit the equations. The condition for the tangent line to both the circle and parabola led us to m⁴ + m² - 2 = 0, which had solutions m² = 1 and m² = -2. Since m² can't be negative, only m = ±1. Therefore, only two common tangents. So that gives two tangent lines, each touching the circle once and the parabola once, hence four points of tangency. Therefore, the quadrilateral formed by these four points is as we calculated, with area 15.

But wait, hold on. Let me check the initial steps again. When we found that the common tangent lines have slopes m = ±1, we found their equations as y = x + 2 and y = -x - 2. However, is there another possibility for common tangents? For example, sometimes there can be more common tangents depending on the curves.

But in this case, since the parabola y² = 8x opens to the right, and the circle is centered at the origin with radius √2 ≈ 1.414, which is smaller than the distance from the origin to the vertex of the parabola (which is at (0,0)), so the circle is entirely inside the parabola? Wait, no. Wait, the parabola y² = 8x extends infinitely to the right, while the circle has a radius of √2. The circle is located at the origin, radius √2 ≈ 1.414. The vertex of the parabola is at (0,0), so the circle is centered at the vertex of the parabola. Therefore, the circle lies partially inside and partially outside the parabola? Let me check.

For example, at x = 0, the circle has points (0, ±√2) ≈ (0, ±1.414). The parabola at x = 0 is y² = 0, so only (0, 0). Therefore, the circle intersects the parabola at the origin? Wait, plugging (0,0) into both equations:

Circle: 0² + 0² = 0 ≠ 2. Therefore, the origin is not on the circle. Therefore, the circle does not pass through the vertex of the parabola. Therefore, the circle is entirely outside the parabola? Or partly inside?

Wait, for example, take a point on the circle, say (x, y) such that x² + y² = 2. Let me see if such a point lies inside or outside the parabola y² = 8x.

The region inside the parabola y² ≤ 8x. For a point (x, y) on the circle, we can check if y² ≤ 8x or not.

But since x² + y² = 2, we can write x = √(2 - y²). Wait, but x must be non-negative because for the parabola y² = 8x, x ≥ 0.

But for the circle, x can be positive or negative. However, for the parabola, x ≥ 0. Therefore, points on the circle with x < 0 are not in the region of the parabola. So only the right half of the circle (x ≥ 0) can be compared to the parabola.

Let me take a point on the circle with x ≥ 0. Let me parametrize it as x = √2 cosθ, y = √2 sinθ. Then, check if y² ≤ 8x.

Compute y² = 2 sin²θ, 8x = 8√2 cosθ.

So the condition is 2 sin²θ ≤ 8√2 cosθ, which is sin²θ ≤ 4√2 cosθ. This is complicated, but let me test specific points.

At θ = 0: x = √2, y = 0. Then, y² = 0 ≤ 8x = 8√2 ≈ 11.31. True.

At θ = π/2: x = 0, y = √2. Then, y² = 2. 8x = 0. 2 ≤ 0? No. Therefore, at θ = π/2, the point (√2 cos(π/2), √2 sin(π/2)) = (0, √2) is not inside the parabola. Similarly, points on the circle with x ≥ 0 may lie inside or outside the parabola depending on θ. Therefore, the circle intersects the parabola?

Wait, let me find if there are intersection points between the circle and parabola.

Solve the system:

x² + y² = 2,

y² = 8x.

Substitute y² from the second equation into the first: x² + 8x = 2 ⇒ x² + 8x - 2 = 0. Solving:

x = [-8 ± sqrt(64 + 8)] / 2 = [-8 ± sqrt(72)] / 2 = [-8 ± 6√2]/2 = -4 ± 3√2.

Approximate sqrt(2) ≈ 1.414, so 3√2 ≈ 4.242.

Therefore, x ≈ -4 + 4.242 ≈ 0.242, or x ≈ -4 - 4.242 ≈ -8.242. Since for the parabola, x ≥ 0, only the positive solution is x ≈ 0.242. Therefore, there is one point of intersection between the circle and the parabola at x ≈ 0.242, y² = 8x ≈ 1.936, so y ≈ ±1.391. Therefore, the circle and parabola intersect at two points. Therefore, they are not entirely separate or one inside the other, but intersect at two points.

Therefore, there can be common tangents. However, we found two common tangents. But the problem states that there are two common tangents, forming four points of tangency. Which are the four points we calculated: two on the circle and two on the parabola.

Therefore, I think our calculation is correct, and the area is 15. However, given that the problem says "the quadrilateral formed by the four points of tangency," and given that the Asymptote code has those coordinates, which give area 15, I think that is the answer.

But just to check once again, let me think if there is a possibility that there are more common tangents? For example, sometimes there can be more than two common tangents between a circle and a parabola. Let me check.

The general condition for a line to be tangent to both curves is what we solved earlier, leading to m = ±1. So only two slopes. Hence only two common tangents.

Alternatively, perhaps there are vertical tangents? For example, vertical lines x = k. Let me check if any vertical lines are tangent to both the circle and the parabola.

For a vertical line x = k to be tangent to the parabola y² = 8x, it must intersect the parabola at exactly one point. Substitute x = k into the parabola equation: y² = 8k. This has real solutions only if 8k ≥ 0 ⇒ k ≥ 0. The line x = k intersects the parabola at (k, ±√(8k)). For this to be a single point, √(8k) must be zero, which happens when k = 0. But x = 0 intersects the parabola only at (0, 0). However, the line x = 0 is tangent to the parabola y² = 8x at (0, 0). Now check if it is tangent to the circle x² + y² = 2. The line x = 0 is the y-axis, which intersects the circle at (0, ±√2). So two points, hence not tangent. Therefore, x = 0 is not a tangent line to the circle. Therefore, no vertical common tangents other than maybe non-existent.

Similarly, horizontal lines y = c. For horizontal lines to be tangent to the parabola y² = 8x, we can check. A horizontal line y = c intersects the parabola at x = y² / 8. For it to be tangent, there should be exactly one x, which is always true, but the line intersects the parabola at only one point if it's tangent? Wait, no. For a horizontal line y = c, substituting into the parabola gives x = c² / 8. So it intersects the parabola at exactly one point (x, c), so actually, all horizontal lines are tangent to the parabola? Wait, no, that can't be. Wait, the parabola y² = 8x is symmetric about the x-axis, and for any horizontal line y = c, there is exactly one x-value (x = c² / 8) where it intersects the parabola. Therefore, every horizontal line is tangent to the parabola? That seems incorrect. Wait, actually, no. Wait, for example, take y = 0. It intersects the parabola at x = 0, which is the vertex. But the line y = 0 is the axis of the parabola and intersects it at one point. However, for y ≠ 0, the horizontal line intersects the parabola at one point. Wait, actually, yes, every horizontal line is tangent to the parabola y² = 8x? No, this is not correct. Because the parabola y² = 8x is not a function; it's symmetric about the x-axis. However, for each horizontal line y = c, it intersects the parabola at exactly one point (x = c²/8, y = c), but that's only one point? Wait, no, it's actually one point? Wait, no, because for each y = c, the equation x = c² / 8. So only one x-value. Therefore, the horizontal line y = c meets the parabola at one point. But the parabola is not a function in terms of x, but it is a function in terms of y, if we solve for y in terms of x, but no. Wait, maybe it's because of the way it's defined. However, in reality, for a parabola, there are infinitely many horizontal lines that intersect it at one point. But in reality, the parabola is not a function, but the horizontal lines intersect it at one point (since for each horizontal line, there is one x-value). However, they are not tangent lines; they are actually secant lines? No, in this case, they are tangent only if they touch at one point. But for the parabola y² = 8x, which opens to the right, the horizontal lines y = c are not tangent lines, because the tangent lines have slopes, and the horizontal lines are just intersecting it at one point, but they are not the tangent lines. Wait, for example, take the line y = 4. Substitute into the parabola: 16 = 8x ⇒ x = 2. So the line y = 4 intersects the parabola at (2, 4). Is that a tangent? To check, we can see if the line is tangent to the parabola. The derivative of the parabola y² = 8x is 2y dy/dx = 8 ⇒ dy/dx = 4/y. At the point (2, 4), dy/dx = 4/4 = 1. Therefore, the tangent line at (2, 4) has slope 1, which is the line y = x + 2, not y = 4. Therefore, the line y = 4 is not tangent to the parabola. Therefore, even though it intersects at one point, it is not a tangent line. Therefore, my previous conclusion was wrong. So horizontal lines are not tangent to the parabola; they just intersect it at one point, but are not tangent.

Therefore, in our case, our only common tangents are y = x + 2 and y = -x - 2. Therefore, only two common tangents, which touch the circle and the parabola at four points, forming the quadrilateral with area 15. Therefore, the answer should be 15.

But I need to check once more if there's an error in my coordinate calculations. Let me verify the shoelace formula once again with different order.

Order: A(-1,1), B(2,4), D(2,-4), C(-1,-1), back to A.

Calculating shoelace sum:

First, multiply x_i * y_{i+1}:

(-1)(4) + 2*(-4) + 2*(-1) + (-1)(1) = (-4) + (-8) + (-2) + (-1) = -15.

Then multiply y_i * x_{i+1}:

(1)(2) + 4*(2) + (-4)*(-1) + (-1)*(-1) = 2 + 8 + 4 + 1 = 15.

Subtract: |-15 - 15| = 30. Area is 15. Correct.

Alternatively, using the formula for the area of a trapezoid. Since the two parallel sides are vertical, the distance between them is the horizontal distance, which is 2 - (-1) = 3. The lengths of the parallel sides are the lengths of the vertical segments:

From (-1,1) to (-1,-1): length is sqrt( (0)^2 + (-1 - 1)^2 ) = sqrt(0 + 4) = 2.

From (2,4) to (2,-4): sqrt( (0)^2 + (-4 - 4)^2 ) = sqrt(0 + 64) = 8.

Then area is ((2 + 8)/2)*3 = 5*3 = 15. Correct.

Therefore, I think confidently that the area is 15.

But just to check once again, let me think of coordinates again. The four points are (-1,1), (2,4), (2,-4), (-1,-1). If I plot these, the quadrilateral is a trapezoid with bases of length 2 and 8, and height 3, which gives area 15. Yep. Therefore, the answer is 15.

Wait, but the Asymptote code draws the graph with those points, and labels A(-1,1), B(2,4), C(-1,-1), D(2,-4). So the problem says "the quadrilateral formed by the four points of tangency." So the answer is 15. Hence, I think that's correct.

**Final Answer**
\boxed{15}

To solve the problem, we are asked to find the area of the quadrilateral formed by the four points of tangency of the circle $ x^2 + y^2 = 2 $ and the parabola $ y^2 = 8x $, which are known to have exactly two common tangents.

---

### Step 1: Identify the Tangent Lines

A **common tangent** is a line that touches both the circle and the parabola at exactly one point each. Let’s find the equations of such lines.

**Equation of the Parabola:**
The general tangent to a parabola of the form $ y^2 = 4ax $ is:

$$
y = mx + \frac{a}{m}
$$

Here, the given parabola is $ y^2 = 8x $, so $ 4a = 8 \Rightarrow a = 2 $.
So, the equation of a line tangent to the parabola is:

$$
y = mx + \frac{2}{m}
$$

---

### Step 2: Use the Condition for Tangency to the Circle

For a line $ y = mx + c $ to also be tangent to the circle $ x^2 + y^2 = 2 $, the perpendicular distance from the center of the circle (which is at $ (0, 0) $) to the line must equal the radius $ \sqrt{2} $.

The perpendicular distance from a point $ (x_0, y_0) $ to a line $ ax + by + c = 0 $ is:

$$
\text{Distance} = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}
$$

For the line $ y = mx + c $, rewrite it as $ mx - y + c = 0 $.
Then, the distance from (0, 0) is:

$$
\frac{|c|}{\sqrt{m^2 + 1}} = \sqrt{2}
\Rightarrow |c| = \sqrt{2} \cdot \sqrt{m^2 + 1}
$$

We already have $ c = \frac{2}{m} $, so:

$$
\frac{2}{m} = \pm \sqrt{2} \cdot \sqrt{m^2 + 1}
\Rightarrow \frac{4}{m^2} = 2(m^2 + 1)
$$

Multiply both sides by $ m^2 $:

$$
4 = 2m^4 + 2m^2
\Rightarrow m^4 + m^2 - 2 = 0
$$

Let $ z = m^2 $, then:

$$
z^2 + z - 2 = 0
\Rightarrow z = \frac{-1 \pm \sqrt{1^2 + 8}}{2} = \frac{-1 \pm 3}{2}
\Rightarrow z = 1 \text{ or } z = -2
$$

Since $ z = m^2 \geq 0 $, we have:

$$
m^2 = 1 \Rightarrow m = \pm 1
$$

---

### Step 3: Find the Tangent Lines and Their Points of Tangency

#### Tangent Line 1: $ m = 1 $

$$
y = x + \frac{2}{1} = x + 2
$$

- This line is **tangent to the parabola** at the point where it intersects it once:

    Substituting $ y = x + 2 $ into $ y^2 = 8x $:

    $$
    (x + 2)^2 = 8x \Rightarrow x^2 + 4x + 4 = 8x \Rightarrow x^2 - 4x + 4 = 0 \Rightarrow (x - 2)^2 = 0 \Rightarrow x = 2
    $$

    Therefore, one tangency point on the parabola is:
    $$
    (2, 4)
    $$

- The tangent point on the **circle** is where this line is tangent:

    We know the radius is perpendicular to the tangent at the point of tangency, and the direction vector of the line is $ (1,1) $, so the radius (from center to point) must be perpendicular. This leads to point:
    $$
    (-1,1)
    $$

#### Tangent Line 2: $ m = -1 $

$$
y = -x - 2
$$

- This line is **tangent to the parabola** at:

    $$
    y^2 = 8x \Rightarrow (-x - 2)^2 = 8x \Rightarrow (x + 2)^2 = 8x \Rightarrow x^2 + 4x + 4 = 8x \Rightarrow x = 2
    \Rightarrow y = -4
    $$

    Tangency point on parabola:
    $$
    (2, -4)
    $$

- The point of tangency on the **circle** is found using perpendicular direction, leading to:
    $$
    (-1, -1)
    $$

---

### Step 4: Find the Area of the Quadrilateral

We now have the four **points of tangency**:
- $ A = (-1, 1) $
- $ B = (2, 4) $
- $ C = (-1, -1) $
- $ D = (2, -4) $

These points form a quadrilateral. Let's order them as $ A, B, D, C $ (assuming this gives a counterclockwise traversal).

We apply the **shoelace formula** to compute the area:

$$
\text{Area} = \frac{1}{2} \left| \sum x_i y_{i+1} - \sum y_i x_{i+1} \right|
$$

Using coordinates:
- $ A = (-1,1), B = (2,4), D = (2,-4), C = (-1,-1) $

Compute:
- $ x_i y_{i+1} $:
  $ (-1)(4) + (2)(-4) + (2)(-1) + (-1)(1) = -4 -8 -2 -1 = -15 $
- $ y_i x_{i+1} $:
  $ (1)(2) + (4)(2) + (-4)(-1) + (-1)(-1) = 2 + 8 + 4 + 1 = 15 $

$$
\text{Area} = \frac{1}{2} \left| -15 - 15 \right| = \frac{1}{2} \times 30 = \boxed{15}
$$

---

### Final Answer

$$
\boxed{15}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of the quadrilateral formed by the four points of tangency between the circle x² + y² = 2 and the parabola y² = 8x. The problem mentions there are two common tangents, forming four points of tangency. First, I need to figure out what these common tangents are, find their points of tangency on both the circle and the parabola, and then determine the quadrilateral they form. Finally, I have to calculate its area.

Let me start by recalling that a common tangent to two curves is a line that touches both curves at exactly one point each. Since the circle and the parabola are given, I need to find equations for these common tangents.

First, let me write down the equations again:

Circle: x² + y² = 2.

Parabola: y² = 8x.

I remember that the general equation of a tangent to a parabola can be written in terms of slope. For the parabola y² = 4ax, the tangent is y = mx + a/m. Comparing this to our parabola y² = 8x, which is equivalent to 4a = 8, so a = 2. Therefore, the tangent to the parabola y² = 8x can be written as y = mx + 2/m. Wait, let me verify that.

Alternatively, for a general parabola y² = 4ax, the equation of tangent with slope m is y = mx + a/m. So yes, in this case, since 4a = 8, a = 2, so the tangent line is y = mx + 2/m. So that's the equation of the tangent line to the parabola with slope m.

Now, this line should also be tangent to the circle x² + y² = 2. The condition for a line y = mx + c to be tangent to the circle x² + y² = r² is that the distance from the center of the circle to the line equals the radius.

The circle here is centered at (0, 0) with radius √2. So the distance from (0, 0) to the line y = mx + c is |0 - m*0 - c| / sqrt(m² + 1) ) = |c| / sqrt(m² + 1). For this to be equal to the radius √2, we have |c| / sqrt(m² + 1) = √2. Therefore, |c| = √2 * sqrt(m² + 1). Therefore, c = ±√2 * sqrt(m² + 1).

But since for the parabola's tangent line, we have c = 2/m. Therefore, setting these equal:

For the line to be tangent to both the parabola and the circle, we must have:

2/m = ±√2 * sqrt(m² + 1)

So this equation can be used to solve for m.

Let me write that equation:

2/m = ±√2 * sqrt(m² + 1)

Let me square both sides to eliminate the square root and the absolute value. But before that, let me note that m cannot be zero because the denominator would be zero. Also, since the parabola y² = 8x opens to the right, the tangent lines might have positive or negative slopes. Let me proceed step by step.

So squaring both sides:

(2/m)² = (√2 * sqrt(m² + 1))²

Which simplifies to:

4/m² = 2*(m² + 1)

Multiply both sides by m²:

4 = 2*(m² + 1)*m²

Divide both sides by 2:

2 = (m² + 1)*m²

So:

m⁴ + m² - 2 = 0

Let me set z = m², then the equation becomes:

z² + z - 2 = 0

Solving this quadratic equation:

z = [-1 ± sqrt(1 + 8)] / 2 = [-1 ± 3]/2

Therefore, z = (-1 + 3)/2 = 2/2 = 1 or z = (-1 - 3)/2 = -4/2 = -2

But since z = m², it can't be negative. Therefore, z = 1, so m² = 1 => m = ±1

Therefore, the slopes are m = 1 and m = -1.

But wait, earlier we had 2/m = ±√2 * sqrt(m² + 1). Let me check if these slopes satisfy the original equation.

Let me check m = 1:

Left side: 2/m = 2/1 = 2

Right side: ±√2 * sqrt(1 + 1) = ±√2 * √2 = ±2

Therefore, 2 = ±2. So the positive sign. So m = 1 gives c = 2/m = 2. Therefore, equation y = x + 2.

Similarly, for m = -1:

Left side: 2/m = 2/(-1) = -2

Right side: ±√2 * sqrt(1 + 1) = ±2. So we need -2 = ±2, which implies that we take the negative sign. Therefore, c = -2. Therefore, the equation is y = -x - 2.

Wait, but let me check if these lines are actually tangent to the parabola. For example, for m = 1, the line is y = x + 2. Let me check if this intersects the parabola y² = 8x.

Substitute y = x + 2 into y² = 8x:

(x + 2)² = 8x => x² + 4x + 4 = 8x => x² - 4x + 4 = 0 => (x - 2)² = 0. Therefore, x = 2, and y = 2 + 2 = 4. So yes, it's tangent at (2, 4). Similarly, for m = -1, the line is y = -x - 2. Substitute into y² = 8x:

(-x - 2)^2 = 8x => x² + 4x + 4 = 8x => x² - 4x + 4 = 0 => (x - 2)^2 = 0. Therefore, x = 2, y = -2 -2 = -4. Therefore, tangent at (2, -4).

Wait, but the problem mentions that there are two common tangents forming four points of tangency. But according to this, each tangent touches the parabola at one point and the circle at one point. So each tangent has two points of tangency? Wait, but the problem says "forming four points of tangency". So maybe there are two tangents, each tangent to both the circle and the parabola, so each tangent touches both curves once, so total four points of tangency? That seems possible. However, the Asymptote code in the problem shows four points A, B, C, D, with A = (-1,1), B = (2,4), C = (-1,-1), D = (2,-4). Wait, but according to my calculation, the tangents touch the parabola at (2,4) and (2,-4). Where are the other points?

Wait, maybe I made a mistake here. Maybe there are more common tangents?

Wait, let me think again. The equations of the common tangents are y = x + 2 and y = -x - 2. However, maybe there are also other common tangents?

Wait, but according to the algebra, solving for m led us to only two possible slopes, m = ±1. However, the Asymptote code shows that there are four points of tangency, with coordinates like (-1,1), (-1,-1), (2,4), (2,-4). So maybe there are two common tangents, each tangent to the circle and parabola, but each tangent is touching the circle at two points? No, a line can only be tangent to a circle at one point. So perhaps there are two common tangents, but each tangent line touches the circle and the parabola once, hence four points of tangency: two on the circle and two on the parabola? Wait, but in that case, the Asymptote code shows that the points of tangency on the parabola are (2,4) and (2,-4), as we found, and the points of tangency on the circle would be... Let me find them.

So for the line y = x + 2, which touches the parabola at (2, 4). Now, let's find the point where this line is tangent to the circle. For that, we need to find the point on the circle where the line is tangent.

Alternatively, since we know that the tangent to the circle is y = x + 2. The tangent line to a circle at a point (x1, y1) is xx1 + yy1 = r². Here, the circle is x² + y² = 2, so the tangent at (x1, y1) is xx1 + yy1 = 2. Let me compare that with the line y = x + 2.

So, let me write the equation of the tangent line to the circle as y = x + 2. Let me rearrange this to standard form: x - y + 2 = 0. Wait, the standard form for the tangent line to the circle is xx1 + yy1 = 2. So equating coefficients?

Alternatively, suppose the tangent line is y = x + 2. Then, since it's tangent to the circle x² + y² = 2, the point of tangency (x1, y1) lies on the line and satisfies the condition that the vector from the center to (x1, y1) is perpendicular to the tangent line.

The slope of the tangent line is 1, so the slope of the radius is -1 (negative reciprocal). Therefore, the line from (0,0) to (x1, y1) has slope y1/x1 = -1, so y1 = -x1. Also, since (x1, y1) lies on the line y = x + 2, substituting y1 = -x1 into that gives -x1 = x1 + 2 → -2x1 = 2 → x1 = -1. Therefore, x1 = -1, y1 = -(-1) = 1. Therefore, the point of tangency on the circle is (-1, 1). Similarly, for the other tangent line y = -x - 2, let's find the point of tangency.

Same process. The tangent line has slope -1, so the radius has slope 1. Therefore, y1/x1 = 1 ⇒ y1 = x1. Since the point lies on the line y = -x - 2, substituting y1 = x1 gives x1 = -x1 - 2 ⇒ 2x1 = -2 ⇒ x1 = -1, so y1 = -1. Therefore, the point of tangency on the circle is (-1, -1). Therefore, the four points of tangency are (-1, 1) on the circle, (2, 4) on the parabola, (-1, -1) on the circle, and (2, -4) on the parabola. Then, these four points form a quadrilateral. The Asymptote code also labels these points as A(-1,1), B(2,4), C(-1,-1), D(2,-4). Then, connecting them as A-B-C-D-A? Wait, how is the quadrilateral formed? The Asymptote code draws lines connecting A to C and B to D. So maybe the quadrilateral is A-B-D-C or A-B-D-C? Wait, looking at the Asymptote code:

draw(interp(A,B,-0.2)--interp(A,B,1.2)); which is the tangent line AB?

Wait, the Asymptote code is a bit unclear, but the key is that the four points of tangency are A(-1,1), B(2,4), C(-1,-1), D(2,-4). So connecting these four points, what quadrilateral do they form? Let me plot them mentally.

Point A(-1,1) is on the left side of the circle, upper half. Point B(2,4) is on the parabola, far right upper. Point C(-1,-1) is on the left side, lower half. Point D(2,-4) is on the parabola, far right lower. So connecting A-B-D-C-A? Let me think. If we connect A to B to D to C to A, that would form a quadrilateral. Alternatively, maybe A-B-C-D? Wait, but C is (-1,-1), so connecting A(-1,1) to B(2,4) to C(-1,-1) would cross over. Maybe it's a kite-shaped figure or some quadrilateral.

Alternatively, since the tangent lines are AB and CD? Wait, the tangent lines are AB: y = x + 2, which touches the parabola at B(2,4) and the circle at A(-1,1). Similarly, the other tangent line is y = -x - 2, which touches the parabola at D(2,-4) and the circle at C(-1,-1). Therefore, the two tangent lines are AB and CD. But how do these points form a quadrilateral? Maybe the quadrilateral is formed by connecting A to B to D to C to A? Let me check coordinates:

A(-1,1), B(2,4), D(2,-4), C(-1,-1). Connecting A to B to D to C to A. Let me visualize:

- From (-1,1) to (2,4): upper right
- From (2,4) to (2,-4): vertical line down the right side
- From (2,-4) to (-1,-1): diagonal left up
- From (-1,-1) back to (-1,1): vertical line up the left side.

Wait, that seems to form a trapezoid? Maybe not. Alternatively, maybe it's a quadrilateral with two sides on the tangent lines and the other sides as connecting the points? Wait, but the problem says "the quadrilateral formed by the four points of tangency", so it's the quadrilateral with vertices at those four points. So regardless of how they're connected, we can compute the area.

But first, let me make sure the four points are correct. So we have:

For the tangent line y = x + 2:

- Point of tangency with circle: (-1, 1)
- Point of tangency with parabola: (2, 4)

For the tangent line y = -x - 2:

- Point of tangency with circle: (-1, -1)
- Point of tangency with parabola: (2, -4)

So these four points: (-1,1), (2,4), (-1,-1), (2,-4). So plotting these, they form a quadrilateral. To find the area, perhaps we can use coordinates to compute the area.

Let me note the coordinates again:

A(-1,1), B(2,4), C(-1,-1), D(2,-4). Wait, but in the Asymptote code, they have A(-1,1), B(2,4), C(-1,-1), D(2,-4). So connecting A, B, D, C? Let me check if that's a quadrilateral. Alternatively, maybe A, B, C, D? But connecting A(-1,1) to B(2,4) to C(-1,-1) to D(2,-4) would cross over. Alternatively, maybe the quadrilateral is A, B, D, C connected as a trapezoid? Let me see.

Alternatively, since there are two tangent lines, each tangent line has two points of tangency. So maybe the quadrilateral is formed by connecting the two points on the circle with the two points on the parabola? But how?

Alternatively, maybe the quadrilateral is a rectangle or some symmetric figure. Let me check the coordinates.

Looking at points A(-1,1), B(2,4), C(-1,-1), D(2,-4). If I think about these points, they are symmetric with respect to both the x-axis and y-axis? Let me check:

Point A(-1,1) and point C(-1,-1) are symmetric over the x-axis.

Point B(2,4) and D(2,-4) are symmetric over the x-axis.

Also, if we consider the y-axis, point A(-1,1) and the point (1,1) would be symmetric, but there is no such point here. Similarly for others.

But perhaps the quadrilateral is symmetric over the x-axis. If that's the case, then maybe I can compute the area of the upper half and double it.

Alternatively, use coordinates to compute the area using the shoelace formula.

Yes, the shoelace formula is a good method for finding the area of a polygon given its coordinates. However, to use that, I need to ensure the points are ordered either clockwise or counterclockwise.

Let me order the points properly. Let's take A(-1,1), B(2,4), D(2,-4), C(-1,-1), and back to A(-1,1). Let me verify if this order is correct.

Plotting the points:

Starting at (-1,1) (A), going to (2,4) (B), then to (2,-4) (D), then to (-1,-1) (C), then back to A. This should create a quadrilateral. Let me check if this is a trapezoid or some other shape. Alternatively, perhaps it's a parallelogram? Let me check the slopes of the sides.

From A(-1,1) to B(2,4): slope is (4 - 1)/(2 - (-1)) = 3/3 = 1.

From B(2,4) to D(2,-4): slope is (-4 - 4)/(2 - 2) = (-8)/0, undefined (vertical line).

From D(2,-4) to C(-1,-1): slope is (-1 - (-4))/(-1 - 2) = (3)/(-3) = -1.

From C(-1,-1) to A(-1,1): slope is (1 - (-1))/(-1 - (-1)) = 2/0, undefined (vertical line).

So the sides are: two sides with slope 1 and -1, and two vertical sides. Therefore, the quadrilateral is a trapezoid with two vertical sides and two non-parallel sides. Wait, but in trapezoid, only one pair of sides is parallel. However, here, the sides AB and CD have slopes 1 and -1, which are negative reciprocals, so they are not parallel. The vertical sides are parallel to each other (both vertical). So yes, this is a trapezoid with the two vertical sides as the parallel sides.

Therefore, the area can be calculated as the average of the lengths of the two parallel sides multiplied by the distance between them.

The two vertical sides are from A(-1,1) to C(-1,-1) and from B(2,4) to D(2,-4). The lengths of these sides:

Left side: from (-1,1) to (-1,-1): length is |1 - (-1)| = 2.

Right side: from (2,4) to (2,-4): length is |4 - (-4)| = 8.

The distance between these two vertical sides is the horizontal distance between x = -1 and x = 2, which is 2 - (-1) = 3.

Therefore, the area would be average of 2 and 8 multiplied by 3: ((2 + 8)/2) * 3 = (10/2) * 3 = 5 * 3 = 15.

Wait, but let me verify this with shoelace formula to make sure.

Alternatively, using shoelace formula for coordinates ordered correctly.

Order of points: A(-1,1), B(2,4), D(2,-4), C(-1,-1), back to A(-1,1).

Shoelace formula:

Area = 1/2 |sum_{i=1 to n} (x_i y_{i+1} - x_{i+1} y_i)|

Let me compute each term:

First, list the coordinates in order:

1. (-1, 1)
2. (2, 4)
3. (2, -4)
4. (-1, -1)
5. (-1, 1) [closing the polygon]

Compute x_i y_{i+1} - x_{i+1} y_i for each i from 1 to 4.

Term 1 (i=1): x1 y2 - x2 y1 = (-1)(4) - (2)(1) = -4 - 2 = -6

Term 2 (i=2): x2 y3 - x3 y2 = (2)(-4) - (2)(4) = -8 - 8 = -16

Term 3 (i=3): x3 y4 - x4 y3 = (2)(-1) - (-1)(-4) = -2 - 4 = -6

Term 4 (i=4): x4 y5 - x5 y4 = (-1)(1) - (-1)(-1) = -1 - 1 = -2

Sum of terms: -6 + (-16) + (-6) + (-2) = -30

Take absolute value and multiply by 1/2: 1/2 * | -30 | = 15. So area is 15. Which matches the previous calculation.

Therefore, the area of the quadrilateral is 15. Wait, but let me check again if the order of points is correct. Because sometimes depending on the order, the shoelace formula can give a wrong area if the points are not ordered correctly.

Let me try another order. Suppose we take A(-1,1), B(2,4), C(-1,-1), D(2,-4), back to A. Wait, would that give a different area?

Let me compute:

Coordinates:

1. (-1,1)
2. (2,4)
3. (-1,-1)
4. (2,-4)
5. (-1,1)

Compute terms:

Term 1: (-1)(4) - (2)(1) = -4 -2 = -6

Term 2: (2)(-1) - (-1)(4) = -2 + 4 = 2

Term 3: (-1)(-4) - (2)(-1) = 4 + 2 = 6

Term 4: (2)(1) - (-1)(-4) = 2 - 4 = -2

Sum: -6 + 2 + 6 - 2 = 0. That can't be right. So that order is not correct. Therefore, the correct order is A, B, D, C.

Alternatively, another order: A(-1,1), C(-1,-1), D(2,-4), B(2,4), back to A. Let me try.

Coordinates:

1. (-1,1)
2. (-1,-1)
3. (2,-4)
4. (2,4)
5. (-1,1)

Compute terms:

Term1: (-1)(-1) - (-1)(1) = 1 +1 = 2

Term2: (-1)(-4) - (2)(-1) = 4 +2 = 6

Term3: (2)(4) - (2)(-4) = 8 + 8 = 16

Term4: (2)(1) - (-1)(4) = 2 +4 = 6

Total sum: 2 +6 +16 +6 = 30. Area is 1/2*|30| = 15. Same result. So regardless of the order as long as it's correct, the area is 15. Therefore, seems like the area is 15.

But let me think again if this makes sense. The quadrilateral has two vertical sides of length 2 and 8, separated by 3 units horizontally. The area of trapezoid is average of the two bases times height. The bases are 2 and 8, height is 3. So yes, (2 + 8)/2 * 3 = 15. Therefore, seems correct.

But let me check with another approach. For example, coordinates of the points:

A(-1,1), B(2,4), C(-1,-1), D(2,-4). Alternatively, we can think of this quadrilateral as two triangles.

For instance, divide the quadrilateral into two triangles: ABC and ACD? Wait, but need to make sure the triangles are non-overlapping.

Alternatively, split into triangle ABD and triangle ACD? Maybe not. Let me think.

Alternatively, since the quadrilateral is symmetric over the x-axis, we can compute the area of the upper half and double it.

The upper half would be the area between points A(-1,1), B(2,4), and the reflection of D over x-axis which is (2,4). Wait, no, maybe not.

Wait, the quadrilateral is symmetric over the x-axis, since if you reflect over x-axis, A(-1,1) goes to C(-1,-1), B(2,4) goes to D(2,-4). So the upper half would be the triangle formed by A, B, and the midpoint?

Alternatively, since the quadrilateral is symmetric, we can compute the area of the upper half (the part above the x-axis) and double it.

The upper half would be the area bounded by A(-1,1), B(2,4), and the x-axis? Wait, not exactly. Wait, the quadrilateral includes points above and below the x-axis. But due to symmetry, maybe the upper half is the area between A, B, and the reflection of C?

Wait, maybe not. Let me instead compute the area of the trapezoid as a combination of two trapezoids or something else. However, since we already used two methods (trapezoid formula and shoelace formula) and both gave 15, which seems consistent, maybe that's the answer.

But just to be thorough, let me check once more using vectors or coordinate geometry.

Alternatively, consider coordinates:

The quadrilateral has vertices at (-1,1), (2,4), (2,-4), (-1,-1). Let me plot them mentally. The figure is a trapezoid with the two vertical sides at x=-1 and x=2. The left vertical side goes from (-1,1) to (-1,-1), which is a length of 2 units. The right vertical side goes from (2,4) to (2,-4), which is a length of 8 units. The horizontal distance between x=-1 and x=2 is 3 units. Then, as per trapezoid area formula, average of the two bases (2 and 8) multiplied by the height (distance between the vertical sides, which is 3). Therefore, area = ((2 + 8)/2)*3 = 5*3 = 15. Yep, same as before.

Alternatively, think of the figure as a rectangle with a certain width and height, but modified.

But since multiple methods give 15, and the coordinates are given in the Asymptote code, which might be a hint as well. The Asymptote code has A(-1,1), B(2,4), C(-1,-1), D(2,-4). So the area is indeed 15.

But let me check again if my calculation of the points of tangency is correct.

For the tangent line y = x + 2:

- On the circle x² + y² = 2. We found the point (-1,1). Let me verify that (-1)^2 + (1)^2 = 1 + 1 = 2, which is correct. So that point is on the circle.

- On the parabola y² = 8x. At point (2,4): 4² = 16, 8*2 = 16. Correct.

Similarly for the line y = -x - 2:

- On the circle: (-1, -1). (-1)^2 + (-1)^2 = 1 + 1 = 2. Correct.

- On the parabola: (2, -4). (-4)^2 = 16, 8*2 = 16. Correct.

Therefore, the points are correct. So the quadrilateral formed by these four points has area 15.

Wait, but the problem says "the quadrilateral formed by the four points of tangency." But according to the Asymptote code, they also drew lines connecting A to C and B to D, which are the vertical lines. That suggests that the quadrilateral is a rectangle? Wait, but according to the coordinates, it's not a rectangle. Because the sides AB and CD have slopes 1 and -1, which are not perpendicular. Wait, let me check angles.

Alternatively, maybe my assumption about the quadrilateral is wrong? Wait, the Asymptote code draws:

draw(interp(A,B,-0.2)--interp(A,B,1.2)); which is the tangent line AB.

Similarly for CD.

Then draw(A--C); draw(B--D); which connects A to C and B to D.

So the figure is formed by the two tangent lines and the two vertical lines connecting the circle points to the parabola points. But the four points are A, B, C, D, forming the quadrilateral. So the sides of the quadrilateral are AB, BD, DC, and CA? Wait, no. If you connect A to B to D to C to A, then the sides are AB, BD, DC, and CA. However, BD is from B(2,4) to D(2,-4), which is vertical line. DC is from D(2,-4) to C(-1,-1). And CA is from C(-1,-1) to A(-1,1). And AB is from A to B.

Wait, but in this case, the quadrilateral is a trapezoid with two vertical sides and two non-parallel sides, which matches our previous calculation.

But just to confirm, if I think of the coordinates, the quadrilateral is symmetric about the x-axis. Also, the area is 15. Let me check if the answer is 15. However, the problem says "the area of the quadrilateral formed by the four points of tangency." So 15 seems correct, but let me check with another method.

Alternatively, consider coordinates matrix:

For shoelace formula, we can write the coordinates in order:

(-1, 1), (2, 4), (2, -4), (-1, -1)

Applying shoelace formula step by step:

List the coordinates:

x: -1, 2, 2, -1, -1

y: 1, 4, -4, -1, 1

Compute sum of x_i y_{i+1}:

(-1)(4) + (2)(-4) + (2)(-1) + (-1)(1) =

(-4) + (-8) + (-2) + (-1) = -15

Compute sum of y_i x_{i+1}:

(1)(2) + (4)(2) + (-4)(-1) + (-1)(-1) =

2 + 8 + 4 + 1 = 15

Then subtract the two sums: |-15 - 15| = |-30| = 30. Then area is 1/2 * 30 = 15. Yep, same result. Therefore, confident that the area is 15.

However, just to be thorough, let me check if there are other common tangents which I might have missed. Earlier, when solving for the slopes, I obtained m = ±1. But is that all?

Let me revisit the equations. The condition for the tangent line to both the circle and parabola led us to m⁴ + m² - 2 = 0, which had solutions m² = 1 and m² = -2. Since m² can't be negative, only m = ±1. Therefore, only two common tangents. So that gives two tangent lines, each touching the circle once and the parabola once, hence four points of tangency. Therefore, the quadrilateral formed by these four points is as we calculated, with area 15.

But wait, hold on. Let me check the initial steps again. When we found that the common tangent lines have slopes m = ±1, we found their equations as y = x + 2 and y = -x - 2. However, is there another possibility for common tangents? For example, sometimes there can be more common tangents depending on the curves.

But in this case, since the parabola y² = 8x opens to the right, and the circle is centered at the origin with radius √2 ≈ 1.414, which is smaller than the distance from the origin to the vertex of the parabola (which is at (0,0)), so the circle is entirely inside the parabola? Wait, no. Wait, the parabola y² = 8x extends infinitely to the right, while the circle has a radius of √2. The circle is located at the origin, radius √2 ≈ 1.414. The vertex of the parabola is at (0,0), so the circle is centered at the vertex of the parabola. Therefore, the circle lies partially inside and partially outside the parabola? Let me check.

For example, at x = 0, the circle has points (0, ±√2) ≈ (0, ±1.414). The parabola at x = 0 is y² = 0, so only (0, 0). Therefore, the circle intersects the parabola at the origin? Wait, plugging (0,0) into both equations:

Circle: 0² + 0² = 0 ≠ 2. Therefore, the origin is not on the circle. Therefore, the circle does not pass through the vertex of the parabola. Therefore, the circle is entirely outside the parabola? Or partly inside?

Wait, for example, take a point on the circle, say (x, y) such that x² + y² = 2. Let me see if such a point lies inside or outside the parabola y² = 8x.

The region inside the parabola y² ≤ 8x. For a point (x, y) on the circle, we can check if y² ≤ 8x or not.

But since x² + y² = 2, we can write x = √(2 - y²). Wait, but x must be non-negative because for the parabola y² = 8x, x ≥ 0.

But for the circle, x can be positive or negative. However, for the parabola, x ≥ 0. Therefore, points on the circle with x < 0 are not in the region of the parabola. So only the right half of the circle (x ≥ 0) can be compared to the parabola.

Let me take a point on the circle with x ≥ 0. Let me parametrize it as x = √2 cosθ, y = √2 sinθ. Then, check if y² ≤ 8x.

Compute y² = 2 sin²θ, 8x = 8√2 cosθ.

So the condition is 2 sin²θ ≤ 8√2 cosθ, which is sin²θ ≤ 4√2 cosθ. This is complicated, but let me test specific points.

At θ = 0: x = √2, y = 0. Then, y² = 0 ≤ 8x = 8√2 ≈ 11.31. True.

At θ = π/2: x = 0, y = √2. Then, y² = 2. 8x = 0. 2 ≤ 0? No. Therefore, at θ = π/2, the point (√2 cos(π/2), √2 sin(π/2)) = (0, √2) is not inside the parabola. Similarly, points on the circle with x ≥ 0 may lie inside or outside the parabola depending on θ. Therefore, the circle intersects the parabola?

Wait, let me find if there are intersection points between the circle and parabola.

Solve the system:

x² + y² = 2,

y² = 8x.

Substitute y² from the second equation into the first: x² + 8x = 2 ⇒ x² + 8x - 2 = 0. Solving:

x = [-8 ± sqrt(64 + 8)] / 2 = [-8 ± sqrt(72)] / 2 = [-8 ± 6√2]/2 = -4 ± 3√2.

Approximate sqrt(2) ≈ 1.414, so 3√2 ≈ 4.242.

Therefore, x ≈ -4 + 4.242 ≈ 0.242, or x ≈ -4 - 4.242 ≈ -8.242. Since for the parabola, x ≥ 0, only the positive solution is x ≈ 0.242. Therefore, there is one point of intersection between the circle and the parabola at x ≈ 0.242, y² = 8x ≈ 1.936, so y ≈ ±1.391. Therefore, the circle and parabola intersect at two points. Therefore, they are not entirely separate or one inside the other, but intersect at two points.

Therefore, there can be common tangents. However, we found two common tangents. But the problem states that there are two common tangents, forming four points of tangency. Which are the four points we calculated: two on the circle and two on the parabola.

Therefore, I think our calculation is correct, and the area is 15. However, given that the problem says "the quadrilateral formed by the four points of tangency," and given that the Asymptote code has those coordinates, which give area 15, I think that is the answer.

But just to check once again, let me think if there is a possibility that there are more common tangents? For example, sometimes there can be more than two common tangents between a circle and a parabola. Let me check.

The general condition for a line to be tangent to both curves is what we solved earlier, leading to m = ±1. So only two slopes. Hence only two common tangents.

Alternatively, perhaps there are vertical tangents? For example, vertical lines x = k. Let me check if any vertical lines are tangent to both the circle and the parabola.

For a vertical line x = k to be tangent to the parabola y² = 8x, it must intersect the parabola at exactly one point. Substitute x = k into the parabola equation: y² = 8k. This has real solutions only if 8k ≥ 0 ⇒ k ≥ 0. The line x = k intersects the parabola at (k, ±√(8k)). For this to be a single point, √(8k) must be zero, which happens when k = 0. But x = 0 intersects the parabola only at (0, 0). However, the line x = 0 is tangent to the parabola y² = 8x at (0, 0). Now check if it is tangent to the circle x² + y² = 2. The line x = 0 is the y-axis, which intersects the circle at (0, ±√2). So two points, hence not tangent. Therefore, x = 0 is not a tangent line to the circle. Therefore, no vertical common tangents other than maybe non-existent.

Similarly, horizontal lines y = c. For horizontal lines to be tangent to the parabola y² = 8x, we can check. A horizontal line y = c intersects the parabola at x = y² / 8. For it to be tangent, there should be exactly one x, which is always true, but the line intersects the parabola at only one point if it's tangent? Wait, no. For a horizontal line y = c, substituting into the parabola gives x = c² / 8. So it intersects the parabola at exactly one point (x, c), so actually, all horizontal lines are tangent to the parabola? Wait, no, that can't be. Wait, the parabola y² = 8x is symmetric about the x-axis, and for any horizontal line y = c, there is exactly one x-value (x = c² / 8) where it intersects the parabola. Therefore, every horizontal line is tangent to the parabola? That seems incorrect. Wait, actually, no. Wait, for example, take y = 0. It intersects the parabola at x = 0, which is the vertex. But the line y = 0 is the axis of the parabola and intersects it at one point. However, for y ≠ 0, the horizontal line intersects the parabola at one point. Wait, actually, yes, every horizontal line is tangent to the parabola y² = 8x? No, this is not correct. Because the parabola y² = 8x is not a function; it's symmetric about the x-axis. However, for each horizontal line y = c, it intersects the parabola at exactly one point (x = c²/8, y = c), but that's only one point? Wait, no, it's actually one point? Wait, no, because for each y = c, the equation x = c² / 8. So only one x-value. Therefore, the horizontal line y = c meets the parabola at one point. But the parabola is not a function in terms of x, but it is a function in terms of y, if we solve for y in terms of x, but no. Wait, maybe it's because of the way it's defined. However, in reality, for a parabola, there are infinitely many horizontal lines that intersect it at one point. But in reality, the parabola is not a function, but the horizontal lines intersect it at one point (since for each horizontal line, there is one x-value). However, they are not tangent lines; they are actually secant lines? No, in this case, they are tangent only if they touch at one point. But for the parabola y² = 8x, which opens to the right, the horizontal lines y = c are not tangent lines, because the tangent lines have slopes, and the horizontal lines are just intersecting it at one point, but they are not the tangent lines. Wait, for example, take the line y = 4. Substitute into the parabola: 16 = 8x ⇒ x = 2. So the line y = 4 intersects the parabola at (2, 4). Is that a tangent? To check, we can see if the line is tangent to the parabola. The derivative of the parabola y² = 8x is 2y dy/dx = 8 ⇒ dy/dx = 4/y. At the point (2, 4), dy/dx = 4/4 = 1. Therefore, the tangent line at (2, 4) has slope 1, which is the line y = x + 2, not y = 4. Therefore, the line y = 4 is not tangent to the parabola. Therefore, even though it intersects at one point, it is not a tangent line. Therefore, my previous conclusion was wrong. So horizontal lines are not tangent to the parabola; they just intersect it at one point, but are not tangent.

Therefore, in our case, our only common tangents are y = x + 2 and y = -x - 2. Therefore, only two common tangents, which touch the circle and the parabola at four points, forming the quadrilateral with area 15. Therefore, the answer should be 15.

But I need to check once more if there's an error in my coordinate calculations. Let me verify the shoelace formula once again with different order.

Order: A(-1,1), B(2,4), D(2,-4), C(-1,-1), back to A.

Calculating shoelace sum:

First, multiply x_i * y_{i+1}:

(-1)(4) + 2*(-4) + 2*(-1) + (-1)(1) = (-4) + (-8) + (-2) + (-1) = -15.

Then multiply y_i * x_{i+1}:

(1)(2) + 4*(2) + (-4)*(-1) + (-1)*(-1) = 2 + 8 + 4 + 1 = 15.

Subtract: |-15 - 15| = 30. Area is 15. Correct.

Alternatively, using the formula for the area of a trapezoid. Since the two parallel sides are vertical, the distance between them is the horizontal distance, which is 2 - (-1) = 3. The lengths of the parallel sides are the lengths of the vertical segments:

From (-1,1) to (-1,-1): length is sqrt( (0)^2 + (-1 - 1)^2 ) = sqrt(0 + 4) = 2.

From (2,4) to (2,-4): sqrt( (0)^2 + (-4 - 4)^2 ) = sqrt(0 + 64) = 8.

Then area is ((2 + 8)/2)*3 = 5*3 = 15. Correct.

Therefore, I think confidently that the area is 15.

But just to check once again, let me think of coordinates again. The four points are (-1,1), (2,4), (2,-4), (-1,-1). If I plot these, the quadrilateral is a trapezoid with bases of length 2 and 8, and height 3, which gives area 15. Yep. Therefore, the answer is 15.

Wait, but the Asymptote code draws the graph with those points, and labels A(-1,1), B(2,4), C(-1,-1), D(2,-4). So the problem says "the quadrilateral formed by the four points of tangency." So the answer is 15. Hence, I think that's correct.

**Final Answer**
\boxed{15}

To solve the problem, we are asked to find the area of the quadrilateral formed by the four points of tangency of the circle $ x^2 + y^2 = 2 $ and the parabola $ y^2 = 8x $, which are known to have exactly two common tangents.

---

### Step 1: Identify the Tangent Lines

A **common tangent** is a line that touches both the circle and the parabola at exactly one point each. Let’s find the equations of such lines.

**Equation of the Parabola:**
The general tangent to a parabola of the form $ y^2 = 4ax $ is:

$$
y = mx + \frac{a}{m}
$$

Here, the given parabola is $ y^2 = 8x $, so $ 4a = 8 \Rightarrow a = 2 $.
So, the equation of a line tangent to the parabola is:

$$
y = mx + \frac{2}{m}
$$

---

### Step 2: Use the Condition for Tangency to the Circle

For a line $ y = mx + c $ to also be tangent to the circle $ x^2 + y^2 = 2 $, the perpendicular distance from the center of the circle (which is at $ (0, 0) $) to the line must equal the radius $ \sqrt{2} $.

The perpendicular distance from a point $ (x_0, y_0) $ to a line $ ax + by + c = 0 $ is:

$$
\text{Distance} = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}
$$

For the line $ y = mx + c $, rewrite it as $ mx - y + c = 0 $.
Then, the distance from (0, 0) is:

$$
\frac{|c|}{\sqrt{m^2 + 1}} = \sqrt{2}
\Rightarrow |c| = \sqrt{2} \cdot \sqrt{m^2 + 1}
$$

We already have $ c = \frac{2}{m} $, so:

$$
\frac{2}{m} = \pm \sqrt{2} \cdot \sqrt{m^2 + 1}
\Rightarrow \frac{4}{m^2} = 2(m^2 + 1)
$$

Multiply both sides by $ m^2 $:

$$
4 = 2m^4 + 2m^2
\Rightarrow m^4 + m^2 - 2 = 0
$$

Let $ z = m^2 $, then:

$$
z^2 + z - 2 = 0
\Rightarrow z = \frac{-1 \pm \sqrt{1^2 + 8}}{2} = \frac{-1 \pm 3}{2}
\Rightarrow z = 1 \text{ or } z = -2
$$

Since $ z = m^2 \geq 0 $, we have:

$$
m^2 = 1 \Rightarrow m = \pm 1
$$

---

### Step 3: Find the Tangent Lines and Their Points of Tangency

#### Tangent Line 1: $ m = 1 $

$$
y = x + \frac{2}{1} = x + 2
$$

- This line is **tangent to the parabola** at the point where it intersects it once:

    Substituting $ y = x + 2 $ into $ y^2 = 8x $:

    $$
    (x + 2)^2 = 8x \Rightarrow x^2 + 4x + 4 = 8x \Rightarrow x^2 - 4x + 4 = 0 \Rightarrow (x - 2)^2 = 0 \Rightarrow x = 2
    $$

    Therefore, one tangency point on the parabola is:
    $$
    (2, 4)
    $$

- The tangent point on the **circle** is where this line is tangent:

    We know the radius is perpendicular to the tangent at the point of tangency, and the direction vector of the line is $ (1,1) $, so the radius (from center to point) must be perpendicular. This leads to point:
    $$
    (-1,1)
    $$

#### Tangent Line 2: $ m = -1 $

$$
y = -x - 2
$$

- This line is **tangent to the parabola** at:

    $$
    y^2 = 8x \Rightarrow (-x - 2)^2 = 8x \Rightarrow (x + 2)^2 = 8x \Rightarrow x^2 + 4x + 4 = 8x \Rightarrow x = 2
    \Rightarrow y = -4
    $$

    Tangency point on parabola:
    $$
    (2, -4)
    $$

- The point of tangency on the **circle** is found using perpendicular direction, leading to:
    $$
    (-1, -1)
    $$

---

### Step 4: Find the Area of the Quadrilateral

We now have the four **points of tangency**:
- $ A = (-1, 1) $
- $ B = (2, 4) $
- $ C = (-1, -1) $
- $ D = (2, -4) $

These points form a quadrilateral. Let's order them as $ A, B, D, C $ (assuming this gives a counterclockwise traversal).

We apply the **shoelace formula** to compute the area:

$$
\text{Area} = \frac{1}{2} \left| \sum x_i y_{i+1} - \sum y_i x_{i+1} \right|
$$

Using coordinates:
- $ A = (-1,1), B = (2,4), D = (2,-4), C = (-1,-1) $

Compute:
- $ x_i y_{i+1} $:
  $ (-1)(4) + (2)(-4) + (2)(-1) + (-1)(1) = -4 -8 -2 -1 = -15 $
- $ y_i x_{i+1} $:
  $ (1)(2) + (4)(2) + (-4)(-1) + (-1)(-1) = 2 + 8 + 4 + 1 = 15 $

$$
\text{Area} = \frac{1}{2} \left| -15 - 15 \right| = \frac{1}{2} \times 30 = \boxed{15}
$$

---

### Final Answer

$$
\boxed{15}
$$

Solution:
Let the equation of a tangent line be $y = mx + b.$

Substituting into the equation $x^2 + y^2 = 2,$ we get
\[x^2 + (mx + b)^2 = 2.\]Then $(m^2 + 1) x^2 + 2bmx + (b^2 - 2) = 0.$  Since we have a tangent, this quadratic has a double root, meaning that its discriminant is 0.  This gives us
\[(2bm)^2 - 4(m^2 + 1)(b^2 - 2) = 0,\]which simplifies to $b^2 = 2m^2 + 2.$

Solving for $x$ in $y = mx + b,$ we get $x = \frac{y - b}{m}.$  Substituting into $y^2 = 8x,$ we get
\[y^2 = \frac{8y - 8b}{m},\]so $my^2 - 8y + 8b = 0.$  Again, the discriminant of this quadratic will also be 0, so
\[64 - 4(m)(8b) = 0.\]Hence, $bm = 2.$

Then $b = \frac{2}{m}.$  Substituting into $b^2 = 2m^2 + 2,$ we get
\[\frac{4}{m^2} = 2m^2 + 2.\]Then $4 = 2m^4 + 2m^2,$ so $m^4 + m^2 - 2 = 0.$  This factors as $(m^2 - 1)(m^2 + 2) = 0.$  Hence, $m^2 = 1,$ so $m = \pm 1.$

If $m = 1,$ then $b = 2.$  If $m = -1,$ then $b = -2.$  Thus, the two tangents are $y = x + 2$ and $y = -x - 2.$

[asy]
unitsize(0.8 cm);

real upperparab (real x) {
  return (sqrt(8*x));
}

real lowerparab (real x) {
  return (-sqrt(8*x));
}

pair A, B, C, D;

A = (-1,1);
B = (2,4);
C = (-1,-1);
D = (2,-4);

draw(graph(upperparab,0,3));
draw(graph(lowerparab,0,3));
draw(Circle((0,0),sqrt(2)));
draw(interp(A,B,-0.2)--interp(A,B,1.2));
draw(interp(C,D,-0.2)--interp(C,D,1.2));
draw(A--C);
draw(B--D);

label("$y = x + 2$", interp(A,B,1.3), NE);
label("$y = -x - 2$", interp(C,D,1.3), SE);

dot(A);
dot(B);
dot(C);
dot(D);
[/asy]

We look at the tangent $y = x + 2.$  Substituting into $x^2 + y^2 = 2,$ we get
\[x^2 + (x + 2)^2 = 2.\]This simplifies to $x^2 + 2x + 1 = (x + 1)^2 = 0,$ so $x = -1.$  Hence, the tangent point on the circle is $(-1,1).$

We have that $x = y - 2.$  Substituting into $y^2 = 8x,$ we get
\[y^2 = 8(y - 2).\]This simplifies to $(y - 4)^2 = 0,$ so $y = 4.$  Hence, the tangent point on the parabola is $(2,4).$

By symmetry, the other two tangent points are $(-1,-1)$ and $(2,-4).$

The quadrilateral in question is a trapezoid with bases 2 and 8, and height 3, so its area is $\frac{2 + 8}{2} \cdot 3 = \boxed{15}.$


Explanation:
Okay, so I need to find the value of n such that the sum of the first n terms of the geometric sequence {1/4, 1/8, 1/16, ...} is equal to 255/512. Hmm, let me think step by step.

First, I remember that the formula for the sum of the first n terms of a geometric sequence is S_n = a_1*(1 - r^n)/(1 - r), where a_1 is the first term and r is the common ratio. Let me verify that. Yes, that seems right. So I need to identify a_1 and r for this sequence.

Looking at the given sequence: the first term a_1 is 1/4. Then the next term is 1/8. To find the common ratio r, I can divide the second term by the first term. So r = (1/8) / (1/4) = (1/8) * (4/1) = 4/8 = 1/2. Let me check the next term to be sure. The third term is 1/16. If we multiply the second term by 1/2, we get (1/8)*(1/2) = 1/16. Yep, so the common ratio r is indeed 1/2.

So now, I can plug these values into the sum formula. The problem states that the sum S_n is 255/512. Let me write that equation:

255/512 = (1/4)*(1 - (1/2)^n)/(1 - 1/2).

Let me simplify the denominator first. 1 - 1/2 is 1/2. So the formula becomes:

255/512 = (1/4)*(1 - (1/2)^n)/(1/2).

Dividing by 1/2 is the same as multiplying by 2, so:

255/512 = (1/4)*2*(1 - (1/2)^n) = (1/2)*(1 - (1/2)^n).

Therefore, simplifying further:

255/512 = (1/2)*(1 - (1/2)^n).

To solve for n, I can multiply both sides by 2:

2*(255/512) = 1 - (1/2)^n.

Calculating the left side: 2*(255)/512 = 510/512. Let me simplify that fraction. Both numerator and denominator are divisible by 2: 510 ÷ 2 = 255, 512 ÷ 2 = 256. So it becomes 255/256.

So now the equation is:

255/256 = 1 - (1/2)^n.

Let me rearrange this equation to solve for (1/2)^n. Subtract 255/256 from both sides:

1 - 255/256 = (1/2)^n.

Calculating the left side: 1 is 256/256, so 256/256 - 255/256 = 1/256.

Therefore:

(1/2)^n = 1/256.

Now, I need to find n such that (1/2)^n equals 1/256. Let me recall that 1/256 is equal to (1/2)^8 because 2^8 is 256. Let me check that. 2^8 = 256. Yes, so (1/2)^8 = 1/256. Therefore, n must be 8.

Wait, but let me double-check my steps to make sure I didn't make a mistake. Starting from the sum formula:

Sum = (1/4)*(1 - (1/2)^n)/(1 - 1/2) = (1/4)*(1 - (1/2)^n)/(1/2) = (1/4)*(2)*(1 - (1/2)^n) = (1/2)*(1 - (1/2)^n). Then setting that equal to 255/512. Then multiplying both sides by 2 gives 510/512 = 1 - (1/2)^n. Simplifying 510/512 to 255/256, so 1 - 255/256 = 1/256 equals (1/2)^n. Which is 1/2^8. Therefore n = 8. That seems to check out.

But let me verify by calculating the sum of the first 8 terms manually, just to be sure. The first term is 1/4, then 1/8, 1/16, 1/32, 1/64, 1/128, 1/256, 1/512. Let me add them up.

Start adding term by term:

Term 1: 1/4 = 0.25

Term 2: 1/8 = 0.125. Total after 2 terms: 0.25 + 0.125 = 0.375

Term 3: 1/16 = 0.0625. Total: 0.375 + 0.0625 = 0.4375

Term 4: 1/32 = 0.03125. Total: 0.4375 + 0.03125 = 0.46875

Term 5: 1/64 = 0.015625. Total: 0.46875 + 0.015625 = 0.484375

Term 6: 1/128 ≈ 0.0078125. Total ≈ 0.484375 + 0.0078125 = 0.4921875

Term 7: 1/256 ≈ 0.00390625. Total ≈ 0.4921875 + 0.00390625 ≈ 0.49609375

Term 8: 1/512 ≈ 0.001953125. Total ≈ 0.49609375 + 0.001953125 ≈ 0.498046875

Now, let me convert 255/512 to decimal to check. 255 divided by 512. Let me compute that. Since 256/512 is 1/2 = 0.5. So 255/512 is 0.5 - 1/512. 1/512 is approximately 0.001953125, so 0.5 - 0.001953125 = 0.498046875. Which matches the sum of the first 8 terms. Therefore, n=8 is correct.

Wait, but let me make sure I didn't make any calculation errors. For example, when I added up the terms, each term is half of the previous term, so the sum should approach 1/4 divided by (1 - 1/2) = 1/4 divided by 1/2 = 1/2. So the total sum as n approaches infinity is 1/2. And 255/512 is approximately 0.498, which is close to 0.5, so 8 terms is reasonable. If I calculated n=8, then the sum is 0.498..., which is 255/512. So that seems correct.

Alternatively, maybe I can solve (1/2)^n = 1/256 algebraically. Let me write both sides with exponents. Since 1/256 is 2^{-8}, and (1/2)^n is 2^{-n}, so 2^{-n} = 2^{-8}, so -n = -8, so n=8. Yep, that's straightforward.

Therefore, all steps check out, so the answer is n=8. Wait, but let me check again the initial equation. Let me recompute S_n with n=8:

Using the formula S_n = a_1*(1 - r^n)/(1 - r). So a_1 is 1/4, r is 1/2, so:

S_8 = (1/4)*(1 - (1/2)^8)/(1 - 1/2) = (1/4)*(1 - 1/256)/(1/2) = (1/4)*(255/256)/(1/2) = (1/4)*(255/256)*2 = (1/2)*(255/256) = 255/512. Which is exactly the given sum. Perfect, that matches. Therefore, n=8 is definitely the correct answer.

But just to be thorough, let me check n=7. If n=7, then S_7 = (1/4)*(1 - (1/2)^7)/(1 - 1/2) = (1/4)*(1 - 1/128)/(1/2) = (1/4)*(127/128)/ (1/2) = (1/4)*(127/128)*2 = (1/2)*(127/128) = 127/256. Let me convert that to decimal. 127 divided by 256. Since 128/256 is 0.5, so 127/256 is 0.5 - 1/256 ≈ 0.5 - 0.00390625 = 0.49609375. Which is less than 255/512 ≈ 0.498046875. Therefore, n=7 gives a smaller sum, and n=8 gives the correct sum. Therefore, n=8 is indeed the answer.

Therefore, after checking multiple ways, I can confidently say that the value of n is 8.

**Final Answer**
The value of $ n $ is \boxed{8}.

To solve the problem, we are given the infinite geometric sequence:

$$
\left\{ \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots \right\}
$$

We are told that the sum of the first $ n $ terms is:

$$
\frac{255}{512}
$$

Our goal is to find $ n $.

---

### Step 1: Identify the Parameters of the Sequence

From the sequence:
- The first term $ a_1 = \frac{1}{4} $
- The common ratio $ r = \frac{1}{2} $, since each term is half the previous one.

---

### Step 2: Apply the Formula for the Sum of a Geometric Sequence

The formula for the sum of the first $ n $ terms of a geometric sequence is:

$$
S_n = a_1 \cdot \frac{1 - r^n}{1 - r}
$$

Substituting the known values:

$$
\frac{255}{512} = \frac{1}{4} \cdot \frac{1 - (1/2)^n}{1 - 1/2}
$$

Simplify the denominator:

$$
\frac{255}{512} = \frac{1}{4} \cdot \frac{1 - (1/2)^n}{1/2}
$$

Since dividing by $ \frac{1}{2} $ is the same as multiplying by 2:

$$
\frac{255}{512} = \frac{1}{4} \cdot 2 \cdot (1 - (1/2)^n)
$$

$$
\frac{255}{512} = \frac{1}{2} \cdot (1 - (1/2)^n)
$$

Now multiply both sides by 2:

$$
\frac{510}{512} = 1 - (1/2)^n
$$

Simplify the left side:

$$
\frac{255}{256} = 1 - (1/2)^n
$$

Subtract 1 from both sides:

$$
- \frac{1}{256} = - (1/2)^n
$$

Multiply both sides by -1:

$$
\frac{1}{256} = (1/2)^n
$$

---

### Step 3: Solve for $ n $

We recognize that:

$$
\frac{1}{256} = \left( \frac{1}{2} \right)^8
$$

So:

$$
(1/2)^n = (1/2)^8 \Rightarrow n = 8
$$

---

### Step 4: Verification (Optional but Recommended)

Let’s calculate the sum of the first 8 terms to confirm:

$$
S_8 = \frac{1}{4} \cdot \left(1 - (1/2)^8\right) / (1 - 1/2) = \frac{1}{4} \cdot \frac{255/256}{1/2} = \frac{1}{4} \cdot \frac{255}{256} \cdot 2 = \frac{255}{512}
$$

This matches the given sum. Therefore, the value of $ n $ is indeed:

$$
\boxed{8}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of n such that the sum of the first n terms of the geometric sequence {1/4, 1/8, 1/16, ...} is equal to 255/512. Hmm, let me think step by step.

First, I remember that the formula for the sum of the first n terms of a geometric sequence is S_n = a_1*(1 - r^n)/(1 - r), where a_1 is the first term and r is the common ratio. Let me verify that. Yes, that seems right. So I need to identify a_1 and r for this sequence.

Looking at the given sequence: the first term a_1 is 1/4. Then the next term is 1/8. To find the common ratio r, I can divide the second term by the first term. So r = (1/8) / (1/4) = (1/8) * (4/1) = 4/8 = 1/2. Let me check the next term to be sure. The third term is 1/16. If we multiply the second term by 1/2, we get (1/8)*(1/2) = 1/16. Yep, so the common ratio r is indeed 1/2.

So now, I can plug these values into the sum formula. The problem states that the sum S_n is 255/512. Let me write that equation:

255/512 = (1/4)*(1 - (1/2)^n)/(1 - 1/2).

Let me simplify the denominator first. 1 - 1/2 is 1/2. So the formula becomes:

255/512 = (1/4)*(1 - (1/2)^n)/(1/2).

Dividing by 1/2 is the same as multiplying by 2, so:

255/512 = (1/4)*2*(1 - (1/2)^n) = (1/2)*(1 - (1/2)^n).

Therefore, simplifying further:

255/512 = (1/2)*(1 - (1/2)^n).

To solve for n, I can multiply both sides by 2:

2*(255/512) = 1 - (1/2)^n.

Calculating the left side: 2*(255)/512 = 510/512. Let me simplify that fraction. Both numerator and denominator are divisible by 2: 510 ÷ 2 = 255, 512 ÷ 2 = 256. So it becomes 255/256.

So now the equation is:

255/256 = 1 - (1/2)^n.

Let me rearrange this equation to solve for (1/2)^n. Subtract 255/256 from both sides:

1 - 255/256 = (1/2)^n.

Calculating the left side: 1 is 256/256, so 256/256 - 255/256 = 1/256.

Therefore:

(1/2)^n = 1/256.

Now, I need to find n such that (1/2)^n equals 1/256. Let me recall that 1/256 is equal to (1/2)^8 because 2^8 is 256. Let me check that. 2^8 = 256. Yes, so (1/2)^8 = 1/256. Therefore, n must be 8.

Wait, but let me double-check my steps to make sure I didn't make a mistake. Starting from the sum formula:

Sum = (1/4)*(1 - (1/2)^n)/(1 - 1/2) = (1/4)*(1 - (1/2)^n)/(1/2) = (1/4)*(2)*(1 - (1/2)^n) = (1/2)*(1 - (1/2)^n). Then setting that equal to 255/512. Then multiplying both sides by 2 gives 510/512 = 1 - (1/2)^n. Simplifying 510/512 to 255/256, so 1 - 255/256 = 1/256 equals (1/2)^n. Which is 1/2^8. Therefore n = 8. That seems to check out.

But let me verify by calculating the sum of the first 8 terms manually, just to be sure. The first term is 1/4, then 1/8, 1/16, 1/32, 1/64, 1/128, 1/256, 1/512. Let me add them up.

Start adding term by term:

Term 1: 1/4 = 0.25

Term 2: 1/8 = 0.125. Total after 2 terms: 0.25 + 0.125 = 0.375

Term 3: 1/16 = 0.0625. Total: 0.375 + 0.0625 = 0.4375

Term 4: 1/32 = 0.03125. Total: 0.4375 + 0.03125 = 0.46875

Term 5: 1/64 = 0.015625. Total: 0.46875 + 0.015625 = 0.484375

Term 6: 1/128 ≈ 0.0078125. Total ≈ 0.484375 + 0.0078125 = 0.4921875

Term 7: 1/256 ≈ 0.00390625. Total ≈ 0.4921875 + 0.00390625 ≈ 0.49609375

Term 8: 1/512 ≈ 0.001953125. Total ≈ 0.49609375 + 0.001953125 ≈ 0.498046875

Now, let me convert 255/512 to decimal to check. 255 divided by 512. Let me compute that. Since 256/512 is 1/2 = 0.5. So 255/512 is 0.5 - 1/512. 1/512 is approximately 0.001953125, so 0.5 - 0.001953125 = 0.498046875. Which matches the sum of the first 8 terms. Therefore, n=8 is correct.

Wait, but let me make sure I didn't make any calculation errors. For example, when I added up the terms, each term is half of the previous term, so the sum should approach 1/4 divided by (1 - 1/2) = 1/4 divided by 1/2 = 1/2. So the total sum as n approaches infinity is 1/2. And 255/512 is approximately 0.498, which is close to 0.5, so 8 terms is reasonable. If I calculated n=8, then the sum is 0.498..., which is 255/512. So that seems correct.

Alternatively, maybe I can solve (1/2)^n = 1/256 algebraically. Let me write both sides with exponents. Since 1/256 is 2^{-8}, and (1/2)^n is 2^{-n}, so 2^{-n} = 2^{-8}, so -n = -8, so n=8. Yep, that's straightforward.

Therefore, all steps check out, so the answer is n=8. Wait, but let me check again the initial equation. Let me recompute S_n with n=8:

Using the formula S_n = a_1*(1 - r^n)/(1 - r). So a_1 is 1/4, r is 1/2, so:

S_8 = (1/4)*(1 - (1/2)^8)/(1 - 1/2) = (1/4)*(1 - 1/256)/(1/2) = (1/4)*(255/256)/(1/2) = (1/4)*(255/256)*2 = (1/2)*(255/256) = 255/512. Which is exactly the given sum. Perfect, that matches. Therefore, n=8 is definitely the correct answer.

But just to be thorough, let me check n=7. If n=7, then S_7 = (1/4)*(1 - (1/2)^7)/(1 - 1/2) = (1/4)*(1 - 1/128)/(1/2) = (1/4)*(127/128)/ (1/2) = (1/4)*(127/128)*2 = (1/2)*(127/128) = 127/256. Let me convert that to decimal. 127 divided by 256. Since 128/256 is 0.5, so 127/256 is 0.5 - 1/256 ≈ 0.5 - 0.00390625 = 0.49609375. Which is less than 255/512 ≈ 0.498046875. Therefore, n=7 gives a smaller sum, and n=8 gives the correct sum. Therefore, n=8 is indeed the answer.

Therefore, after checking multiple ways, I can confidently say that the value of n is 8.

**Final Answer**
The value of $ n $ is \boxed{8}.

To solve the problem, we are given the infinite geometric sequence:

$$
\left\{ \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots \right\}
$$

We are told that the sum of the first $ n $ terms is:

$$
\frac{255}{512}
$$

Our goal is to find $ n $.

---

### Step 1: Identify the Parameters of the Sequence

From the sequence:
- The first term $ a_1 = \frac{1}{4} $
- The common ratio $ r = \frac{1}{2} $, since each term is half the previous one.

---

### Step 2: Apply the Formula for the Sum of a Geometric Sequence

The formula for the sum of the first $ n $ terms of a geometric sequence is:

$$
S_n = a_1 \cdot \frac{1 - r^n}{1 - r}
$$

Substituting the known values:

$$
\frac{255}{512} = \frac{1}{4} \cdot \frac{1 - (1/2)^n}{1 - 1/2}
$$

Simplify the denominator:

$$
\frac{255}{512} = \frac{1}{4} \cdot \frac{1 - (1/2)^n}{1/2}
$$

Since dividing by $ \frac{1}{2} $ is the same as multiplying by 2:

$$
\frac{255}{512} = \frac{1}{4} \cdot 2 \cdot (1 - (1/2)^n)
$$

$$
\frac{255}{512} = \frac{1}{2} \cdot (1 - (1/2)^n)
$$

Now multiply both sides by 2:

$$
\frac{510}{512} = 1 - (1/2)^n
$$

Simplify the left side:

$$
\frac{255}{256} = 1 - (1/2)^n
$$

Subtract 1 from both sides:

$$
- \frac{1}{256} = - (1/2)^n
$$

Multiply both sides by -1:

$$
\frac{1}{256} = (1/2)^n
$$

---

### Step 3: Solve for $ n $

We recognize that:

$$
\frac{1}{256} = \left( \frac{1}{2} \right)^8
$$

So:

$$
(1/2)^n = (1/2)^8 \Rightarrow n = 8
$$

---

### Step 4: Verification (Optional but Recommended)

Let’s calculate the sum of the first 8 terms to confirm:

$$
S_8 = \frac{1}{4} \cdot \left(1 - (1/2)^8\right) / (1 - 1/2) = \frac{1}{4} \cdot \frac{255/256}{1/2} = \frac{1}{4} \cdot \frac{255}{256} \cdot 2 = \frac{255}{512}
$$

This matches the given sum. Therefore, the value of $ n $ is indeed:

$$
\boxed{8}
$$

Solution:
This is a geometric sequence with first term $\frac{1}{4}$ and common ratio $\frac{1}{2}$. Thus the sum of the first $n$ terms is:

$\frac{255}{512}=\frac{1}{4}\left(\frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}}\right)=\frac{2^n-1}{2^{n+1}}$.

We see that $\frac{255}{512}=\frac{2^8-1}{2^9}$, so $n=\boxed{8}$.