<Question>
The circle $x^2 + y^2 = 2$ and the parabola $y^2 = 8x$ have two common tangents, forming four points of tangency.  Find the area of the quadrilateral formed by the four points of tangency.

[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);

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

<think>
<Plan_1>
Okay, so I need to find the area of the quadrilateral formed by the four points where the circle x² + y² = 2 and the parabola y² = 8x have their common tangents. Hmm, let me start by understanding the problem better.

First, let me recall what the circle and parabola look like. The circle x² + y² = 2 is centered at the origin with radius √2. The parabola y² = 8x opens to the right, with its vertex at the origin. Since the coefficient is 8, it's a bit wider than the standard parabola y² = 4ax, which would have a focus at (a,0). Here, 4a = 8, so a = 2. So the focus of the parabola is at (2,0).

The Asymptote code provided draws the parabola with some points: upperparab (real x) returns sqrt(8x) and lowerparab returns -sqrt(8x). So, for x from 0 to 3, it's plotting the upper and lower halves of the parabola. The circle is drawn with radius sqrt(2), which is approximately 1.414.

Looking at the Asymptote code, the points of tangency are given as A(-1,1), B(2,4), C(-1,-1), D(2,-4). Wait, but how are these points obtained? Let me check if these points lie on both the circle and the parabola.

<prompt_1.1> Verify if the points A(-1,1), B(2,4), C(-1,-1), and D(2,-4) lie on both the circle and the parabola. </prompt_1.1>

The quadrilateral formed by these four points is a kite, since two points are at (-1,1) and (-1,-1), which are vertically aligned, and the other two points are at (2,4) and (2,-4), which are also vertically aligned. But wait, connecting these points: A(-1,1) to B(2,4) to C(-1,-1) to D(2,-4) and back to A? Wait, actually, looking at the Asymptote code, it draws A to C and B to D, which are the diagonals. So the quadrilateral is actually formed by the two tangents (lines AC and BD) and the points where they touch the circle and parabola?

Wait, maybe I need to visualize this better. The common tangents to both the circle and the parabola will each touch the circle at one point and the parabola at another point. Since there are two common tangents, each tangent will have two points of tangency: one on the circle and one on the parabola. Therefore, the four points of tangency (two per tangent) form a quadrilateral.

<prompt_1.2> Explain how the common tangents form a quadrilateral from the points of tangency on the circle and parabola. </prompt_1.2>

But how do I find the equations of these common tangents? Let me recall that a tangent to the parabola y² = 8x can be written in the form y = mx + c, where c = 2/m. Because for the parabola y² = 4ax, the tangent in slope form is y = mx + a/m. Here, 4a = 8, so a = 2. Therefore, the tangent is y = mx + 2/m. So, for any slope m, the tangent to the parabola is y = mx + 2/m.

Similarly, the tangent to the circle x² + y² = 2 can be written as y = mx + c, where the condition for tangency is that the distance from the center (0,0) to the line is equal to the radius. The distance from (0,0) to the line mx - y + c = 0 is |c| / sqrt(m² + 1). The radius is √2, so |c| / sqrt(m² + 1) = √2. Therefore, |c| = √2 * sqrt(m² + 1). So, c = ±√2 * sqrt(m² + 1).

<prompt_1.3> Derive the conditions for a line to be tangent to both the parabola y² = 8x and the circle x² + y² = 2, and find the common tangents. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
For example, if x = 2 on the parabola, y² = 16, so y = ±4. So (2,4) and (2,-4) are on the parabola. Let's check if they are on the circle: x² + y² = 4 + 16 = 20, which is not equal to 2. Wait, that's a problem. The points given in the Asymptote code are A(-1,1), B(2,4), C(-1,-1), D(2,-4). But (2,4) is on the parabola but not on the circle. Hmm, maybe the Asymptote code is just an illustration, not to scale? Or perhaps those points are points of tangency, but not necessarily lying on both the circle and parabola. Wait, no. The points of tangency should lie on both the circle and the parabola.
</execution_1.1>

<execution_1.2>
The Asymptote code draws the circle and parabola, and then two lines (interp(A,B,-0.2)--interp(A,B,1.2) and similarly for C and D). These lines are the common tangents, each tangent to both the circle and the parabola. Each tangent line touches the circle at one point (either A or C) and the parabola at another point (either B or D). Therefore, the four points of tangency (A, B, C, D) form a quadrilateral.
</execution_1.2>

<execution_1.3>
Therefore, the tangent to the circle is y = mx ± √2 * sqrt(m² + 1).

But since the line is a common tangent to both the circle and the parabola, it must satisfy both conditions. That is, the same m and c must satisfy both equations. For the parabola, c = 2/m, and for the circle, |c| = √2 * sqrt(m² + 1). Therefore, substituting c from the parabola's tangent into the circle's condition:

|2/m| = √2 * sqrt(m² + 1)
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.3:
Since c can be positive or negative, but since the parabola y² = 8x is only in the right half-plane (x ≥ 0), the common tangents must also touch the parabola there. Therefore, the y-intercept c can be positive or negative, corresponding to upper and lower tangents. So, we can drop the absolute value by squaring both sides:

(2/m)² = 2(m² + 1)

<prompt_2.1> Solve for the slopes of the common tangents to both the circle and the parabola. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
So, 4/m² = 2(m² + 1)

Multiply both sides by m²:

4 = 2(m² + 1)m²

Divide both sides by 2:

2 = (m² + 1)m²

Expand the right-hand side:

2 = m⁴ + m²

Bring all terms to one side:

m⁴ + m² - 2 = 0

Let me set u = m². Then the equation becomes:

u² + u - 2 = 0

Solving for u:

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

So, u = (-1 + 3)/2 = 1 or u = (-1 - 3)/2 = -2

But u = m² cannot be negative, so u = 1. Therefore, m² = 1 => m = ±1

Therefore, the slopes of the common tangents are m = 1 and m = -1.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Therefore, the equations of the common tangents are:

For m = 1:

<prompt_3.1> Find the equations of the common tangents with slopes m = 1 and m = -1. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
y = 1*x + 2/1 = x + 2

But wait, the parabola's tangent is y = mx + 2/m. For m = 1, that's y = x + 2. But let's check the distance from the circle's center to this line. The distance should be √2.

The line is x - y + 2 = 0. The distance from (0,0) is |0 - 0 + 2| / sqrt(1 + 1) = 2 / sqrt(2) = sqrt(2). Yes, that's correct. So this line is tangent to both the circle and the parabola.

Similarly, for m = -1:

The tangent line is y = -1*x + 2/(-1) = -x - 2

Again, check the distance from the center to this line: |-0 -0 -2| / sqrt(1 + 1) = 2 / sqrt(2) = sqrt(2). Correct. So this line is also a common tangent.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But wait, if these are the only two common tangents, then there are two lines, each with two points of tangency: one on the circle and one on the parabola. Therefore, each tangent line will touch the circle at one point and the parabola at another point. So, in total, four points: two on the circle and two on the parabola? Wait, no. Each tangent line touches both the circle and the parabola once each. So, two tangent lines would give two points on the circle and two points on the parabola. But the problem states "four points of tangency," so maybe each tangent line is considered to have two points (one on each curve), so two lines give four points total? That must be the case.
Therefore, the four points are the two points where the first tangent line touches the circle and parabola, and the two points where the second tangent line touches the circle and parabola.

Therefore, the quadrilateral is formed by these four points: two on the circle and two on the parabola, each pair connected by a tangent line.

Therefore, to find the area of the quadrilateral, I need to find the coordinates of these four points, then compute the area of the quadrilateral they form.

<prompt_4.1> Determine the points of tangency for the common tangents found. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Let's proceed step by step.

First, find the points where the tangent lines touch the circle and the parabola.

Starting with the first tangent line: y = x + 2.

Find the point of tangency on the circle x² + y² = 2. Since the line is tangent to the circle, there's exactly one solution to the system. Let's substitute y = x + 2 into the circle equation:

x² + (x + 2)² = 2

Expand:

x² + x² + 4x + 4 = 2

Combine like terms:

2x² + 4x + 4 - 2 = 0 => 2x² + 4x + 2 = 0

Divide by 2:

x² + 2x + 1 = 0 => (x + 1)^2 = 0 => x = -1

Therefore, x = -1, then y = (-1) + 2 = 1. So the point of tangency on the circle is (-1, 1).
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Similarly, find the point of tangency on the parabola y² = 8x. Substitute y = x + 2 into the parabola equation:

<prompt_5.1> Find the points of tangency for the other common tangent line. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
(y)^2 = 8x

But y = x + 2, so:

(x + 2)^2 = 8x

Expand:

x² + 4x + 4 = 8x

Bring all terms to left:

x² + 4x + 4 - 8x = x² - 4x + 4 = 0

Factor:

(x - 2)^2 = 0 => x = 2

Then y = 2 + 2 = 4. So the point of tangency on the parabola is (2, 4).

Now, for the second tangent line: y = -x - 2.

Find the point of tangency on the circle x² + y² = 2. Substitute y = -x - 2 into the circle equation:

x² + (-x - 2)^2 = 2

Expand:

x² + x² + 4x + 4 = 2

Same as before:

2x² + 4x + 4 = 2 => 2x² + 4x + 2 = 0 => x² + 2x + 1 = 0 => (x + 1)^2 = 0 => x = -1

Then y = -(-1) - 2 = 1 - 2 = -1. So the point of tangency on the circle is (-1, -1).
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Similarly, find the point of tangency on the parabola y² = 8x. Substitute y = -x - 2 into the parabola equation:

<prompt_6.1> Verify the calculations for the points of tangency on the parabola for the second tangent line. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
(y)^2 = 8x

But y = -x - 2, so:

(-x - 2)^2 = 8x

Expand:

x² + 4x + 4 = 8x

Bring all terms to left:

x² + 4x + 4 - 8x = x² - 4x + 4 = 0 => (x - 2)^2 = 0 => x = 2

Then y = -2 - 2 = -4. So the point of tangency on the parabola is (2, -4).

Therefore, the four points of tangency are:

From the first tangent line (y = x + 2):

- On the circle: (-1, 1)
- On the parabola: (2, 4)

From the second tangent line (y = -x - 2):

- On the circle: (-1, -1)
- On the parabola: (2, -4)

So these four points form the quadrilateral. The Asymptote code in the problem draws the circle, parabola, and the two tangent lines, which connect these four points. The quadrilateral is formed by connecting (-1,1) to (2,4), then to (-1,-1), then to (2,-4), and back to (-1,1). Wait, but actually, looking at the coordinates, if you plot them, (-1,1) and (-1,-1) are vertical points, and (2,4) and (2,-4) are vertical points as well. So the quadrilateral is actually a kite, with two vertical lines at x = -1 and x = 2, connected by the points (1 and -1 in y) and (4 and -4 in y).

But to compute the area, perhaps we can use coordinates. Let's list the four points in order. Let me arrange them in a cyclic order. Let's take them as A(-1,1), B(2,4), C(-1,-1), D(2,-4). So connecting A to B to C to D and back to A. Wait, but this might not form a simple quadrilateral. Alternatively, maybe the quadrilateral is formed by the two tangent lines and the points where they touch the circle and parabola. Wait, but each tangent line touches both the circle and the parabola at one point each, so each tangent line is associated with two points (one on the circle, one on the parabola). Therefore, two tangent lines give four points total: two on the circle and two on the parabola. But the problem states "four points of tangency," which would be the four points where the two tangent lines touch the two curves. Each tangent line touches each curve once, so two lines give two points per curve, total four points. Therefore, the quadrilateral is formed by these four points.

But how exactly are these four points connected? The Asymptote code draws the two tangent lines, each connecting a point on the circle to a point on the parabola. The points on the circle are (-1,1) and (-1,-1), and the points on the parabola are (2,4) and (2,-4). So each tangent line connects a circle point to a parabola point. The upper tangent line connects (-1,1) to (2,4), and the lower tangent line connects (-1,-1) to (2,-4). Therefore, the quadrilateral is formed by the four points: (-1,1), (2,4), (-1,-1), (2,-4). So the quadrilateral is a kite-shaped figure with vertices at these four points. To find its area, we can use the shoelace formula.

Alternatively, since it's a kite, the area can be calculated as the sum of the areas of the two triangles formed by the diagonals. The diagonals are the lines connecting (-1,1) to (-1,-1) and (2,4) to (2,-4). Wait, but these diagonals are vertical lines. The first diagonal is from (-1,1) to (-1,-1), which has length 2 (from y=1 to y=-1). The second diagonal is from (2,4) to (2,-4), which has length 8 (from y=4 to y=-4). The area of a kite is (d1 * d2)/2, where d1 and d2 are the lengths of the diagonals. Therefore, area = (2 * 8)/2 = 8. But wait, that seems too straightforward. However, the diagonals in a kite are the ones connecting the two distinct pairs of vertices. Wait, but in this case, the quadrilateral is actually a rectangle? Wait, no. Because the diagonals are the vertical lines at x = -1 and x = 2. Wait, but if the quadrilateral has vertices at (-1,1), (2,4), (-1,-1), (2,-4), then connecting them in order would form a four-sided figure where two sides are vertical and two sides are slanted. Wait, no. Let me plot the points:

- Point A: (-1,1)
- Point B: (2,4)
- Point C: (-1,-1)
- Point D: (2,-4)

If we connect A to B to C to D to A, the sides are AB, BC, CD, DA. Let's compute the lengths:

AB: from (-1,1) to (2,4). Distance is sqrt[(2 - (-1))² + (4 - 1)²] = sqrt[9 + 9] = sqrt[18] = 3√2.

BC: from (2,4) to (-1,-1). Distance is sqrt[(-1 - 2)² + (-1 - 4)²] = sqrt[9 + 25] = sqrt[34].

CD: from (-1,-1) to (2,-4). Distance is sqrt[(2 - (-1))² + (-4 - (-1))²] = sqrt[9 + 9] = sqrt[18] = 3√2.

DA: from (2,-4) to (-1,1). Distance is sqrt[(-1 - 2)² + (1 - (-4))²] = sqrt[9 + 25] = sqrt[34].

So the sides are 3√2, sqrt(34), 3√2, sqrt(34). Therefore, the quadrilateral is a kite with two pairs of adjacent sides equal. The diagonals are AC and BD.

Wait, but in the previous thought, I considered the diagonals as the vertical lines x=-1 and x=2, but actually, in a kite, the diagonals are the lines connecting the vertices. Let me clarify.

If the quadrilateral is A(-1,1), B(2,4), C(-1,-1), D(2,-4), then the diagonals are AC and BD. Wait, connecting A to C and B to D. AC is from (-1,1) to (-1,-1), which is vertical, length 2. BD is from (2,4) to (2,-4), which is vertical, length 8. Then the area would be (d1 * d2)/2 = (2 * 8)/2 = 8. However, this would be the case if the diagonals are perpendicular. But in this case, the diagonals are both vertical, so they are parallel, not perpendicular. Therefore, the formula for the area of a kite (or a quadrilateral with perpendicular diagonals) doesn't apply here. So that approach is incorrect.

Therefore, I need another method. Let's list the coordinates again:

A(-1,1), B(2,4), C(-1,-1), D(2,-4). Let's arrange them in order to form a quadrilateral. If we connect A to B to D to C and back to A, that might form a convex quadrilateral. Wait, but depending on the order, the area can change. Let me make sure of the order.

Looking at the Asymptote code, it draws interp(A,B,-0.2)--interp(A,B,1.2) and similarly for C and D. So the tangents are drawn extending beyond the points A,B and C,D. The quadrilateral is formed by the four points A,B,C,D. So the order is A connected to B, B connected to D, D connected to C, C connected to A? Wait, the Asymptote code draws interp(A,B,-0.2)--interp(A,B,1.2), which is the tangent line AB, and similarly CD. But the quadrilateral is probably the one formed by the four points, connected in the order A,B,D,C or something else. Wait, perhaps the quadrilateral is a bowtie shape? But since the points are two on the left (A and C) and two on the right (B and D), the quadrilateral is formed by the two upper and lower points connected via the tangents. Wait, maybe the quadrilateral is actually a rectangle? But the coordinates don't suggest that.

Alternatively, perhaps the quadrilateral is a trapezoid. Let's check the slopes of the sides.

Slope of AB: (4 -1)/(2 - (-1)) = 3/3 = 1

Slope of BC: (-1 -4)/(-1 -2) = (-5)/(-3) = 5/3

Slope of CD: (-4 - (-1))/(2 - (-1)) = (-3)/3 = -1

Slope of DA: (1 - (-4))/(-1 -2) = 5/(-3) = -5/3

So the sides have slopes 1, 5/3, -1, -5/3. So opposite sides have slopes that are negative reciprocals? Wait, 1 and -1 are not negative reciprocals. 5/3 and -5/3 are negative of each other, but not negative reciprocals. So it's not a trapezoid. Therefore, it's a convex quadrilateral with two sides of slope 1 and -1, and the other two sides with slopes 5/3 and -5/3. Hmm.

Alternatively, maybe the quadrilateral is a parallelogram? Let's check if opposite sides are equal and parallel. From earlier, AB and CD have length 3√2 and slopes 1 and -1. BC and DA have length sqrt(34) and slopes 5/3 and -5/3. Since the opposite sides are not equal in length (3√2 vs sqrt(34)), it's not a parallelogram.

Therefore, to find the area, perhaps the best way is to use the shoelace formula. Let's list the coordinates in order. The order is important; we need to list them either clockwise or counter-clockwise. Let's arrange the points in order as A(-1,1), B(2,4), D(2,-4), C(-1,-1), and back to A. Let's verify:

A(-1,1) to B(2,4): moving right and up.

B(2,4) to D(2,-4): moving straight down.

D(2,-4) to C(-1,-1): moving left and up.

C(-1,-1) to A(-1,1): moving up.

This forms a quadrilateral. Let's apply the shoelace formula.

Shoelace formula: For points (x1,y1), (x2,y2),...., (xn,yn), (x1,y1), the area is |1/2 * sum_{i=1 to n} (xi*yi+1 - xi+1*yi)|

So let's list the coordinates in order: A(-1,1), B(2,4), D(2,-4), C(-1,-1), A(-1,1).

Compute the terms:

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

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

x3*y4 - x4*y3: 2*(-1) - (-1)*(-4) = -2 - 4 = -6

x4*y1 - x1*y4: (-1)*1 - (-1)*(-1) = -1 -1 = -2

Sum these up: -6 + (-16) + (-6) + (-2) = -26

Take absolute value and divide by 2: | -26 | /2 = 13

Therefore, the area is 13.

Wait, but let me verify the order. If I take the points in the order A, B, D, C, the shoelace formula gives 13. However, if I take a different order, the result might be the same in absolute value. Let's check another order to confirm.

Alternatively, order could be A, B, C, D. Let's try that:

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

Compute terms:

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

x2*y3 - x3*y2: 2*(-1) - (-1)*4 = -2 +4 = 2

x3*y4 - x4*y3: (-1)*(-4) - 2*(-1) = 4 +2 =6

x4*y1 - x1*y4: 2*1 - (-1)*(-4) = 2 -4 = -2

Sum: -6 +2 +6 -2 = 0. That can't be right. So the order A,B,C,D is not correct. Therefore, the correct order must be A,B,D,C as before, giving area 13.

Alternatively, maybe the quadrilateral is A,B,C,D ordered differently. Let me check another possible order. Suppose we take A(-1,1), C(-1,-1), D(2,-4), B(2,4), back to A. Let's compute:

x1*y2 - x2*y1: (-1)*(-1) - (-1)*1 = 1 +1 =2

x2*y3 - x3*y2: (-1)*(-4) -2*(-1) =4 +2=6

x3*y4 - x4*y3:2*4 -2*(-4)=8 +8=16

x4*y1 -x1*y4:2*1 - (-1)*4=2 +4=6

Sum:2 +6 +16 +6=28. Absolute value /2=14. Hmm, different result. So which order is correct?

Wait, perhaps the order is not consistent. Let me think. The quadrilateral is formed by two tangent lines. Each tangent line connects a point on the circle to a point on the parabola. The upper tangent line connects (-1,1) to (2,4), and the lower tangent line connects (-1,-1) to (2,-4). Therefore, the quadrilateral is formed by the four points connected as (-1,1) -> (2,4) -> (2,-4) -> (-1,-1) -> (-1,1). That forms a quadrilateral that is symmetric with respect to the x-axis. Therefore, the correct order is A,B,D,C. So the shoelace formula with that order gives 13. However, when I took the order A,C,D,B, it gave 14, which is different.

Wait, perhaps I made a calculation error in the second order. Let me recheck:

Order A,C,D,B:

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

Compute terms:

x1*y2 - x2*y1: (-1)*(-1) - (-1)*1 = 1 - (-1) = 2

x2*y3 - x3*y2: (-1)*(-4) -2*(-1) = 4 - (-2) =6

x3*y4 - x4*y3:2*4 -2*(-4)=8 - (-8)=16

x4*y1 -x1*y4:2*1 - (-1)*4=2 - (-4)=6

Sum:2 +6 +16 +6=28. Absolute value /2=14. Hmm, discrepancy here. Which order is correct?

The correct order should follow the sequence of the quadrilateral's edges. Since the quadrilateral is formed by the two tangent lines, each connecting a circle point to a parabola point, the edges are AB (upper tangent), BD (lower tangent), DC (reverse of upper tangent), and CA (reverse of lower tangent). Wait, perhaps the correct order is A,B,D,C. Let's see:

A(-1,1) connected to B(2,4), then B connected to D(2,-4), then D connected to C(-1,-1), then C connected back to A. So the order is A,B,D,C. Let's compute shoelace formula with this order:

Coordinates:

1. (-1,1)
2. (2,4)
3. (2,-4)
4. (-1,-1)
5. (-1,1) [back to start]

Compute terms:

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

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

x3*y4 - x4*y3: 2*(-1) - (-1)*(-4) = -2 -4 = -6

x4*y1 - x1*y4: (-1)*1 - (-1)*(-1) = -1 -1 = -2

Sum: -6 -16 -6 -2 = -26. Absolute value 26, divided by 2 is 13. So that's the same as before. Therefore, the correct order is A,B,D,C, giving area 13. The other order A,C,D,B is incorrect because it does not follow the actual connectivity of the quadrilateral.

Therefore, the area is 13.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
But wait, let me confirm with another method. Since the quadrilateral is symmetric with respect to the x-axis, we can compute the area of the upper half and double it.

The upper half consists of the points A(-1,1) and B(2,4), connected to the circle and parabola. Wait, actually, the upper half would be the region between the upper tangent line and the x-axis. But perhaps decomposing the quadrilateral into two congruent parts.

Alternatively, since the quadrilateral is symmetric over the x-axis, the area can be calculated as twice the area of the upper half. The upper half would be a quadrilateral with vertices at A(-1,1), B(2,4), and the reflection of C and D? Wait, maybe not. Alternatively, the upper part is a triangle and the lower part is another triangle, but given the coordinates, perhaps integrating or using coordinates is better.

Alternatively, since the quadrilateral is symmetric about the x-axis, the area can be computed as twice the area of the upper half. The upper half is a polygon with vertices at A(-1,1), B(2,4), and the reflection across the x-axis would be A'(-1,-1), B'(2,-4). But the lower half is C and D. Wait, perhaps not. Alternatively, since the quadrilateral is symmetric, we can compute the area of the upper half and double it.

The upper half would be a triangle formed by points A, B, and the intersection of AB with the x-axis. Wait, but the quadrilateral is a four-sided figure. Alternatively, the upper half is a trapezoid between the upper tangent line and the x-axis. Wait, perhaps not. Let me think differently.

Alternatively, since the quadrilateral is symmetric, split it into two triangles across the x-axis. The upper triangle is formed by points A, B, and the projection onto the x-axis. Wait, perhaps not. Alternatively, compute the area using coordinates.

Alternatively, use vectors or coordinates. But the shoelace formula already gave 13, which seems correct. Let me verify once more:

Shoelace formula with points in order A,B,D,C:

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

Compute the terms:

First term: x_A * y_B - x_B * y_A = (-1)*4 - 2*1 = -4 -2 = -6

Second term: x_B * y_D - x_D * y_B = 2*(-4) -2*4 = -8 -8 = -16

Third term: x_D * y_C - x_C * y_D = 2*(-1) - (-1)*(-4) = -2 -4 = -6

Fourth term: x_C * y_A - x_A * y_C = (-1)*1 - (-1)*(-1) = -1 -1 = -2

Sum: -6 -16 -6 -2 = -26. Absolute value 26, divided by 2 gives 13.
</Plan_7>

<Execution_7>
No prompt to execute.
</Execution_7>

<Final_answer>
Therefore, the final answer is $\boxed{13}$.
</Final_answer>
</think>