### Instruction: The parabolas $y^2 = 3x$ and $y^2 = -3x$ each intersect the circle $x^2 + y^2 = 4$ at two points, giving four points of intersection total.  Find the area of the quadrilateral whose vertices are these four points.### Response: To find the area of the quadrilateral, I need to know the coordinates of its vertices.
I can find the coordinates by solving the system of equations that represents the intersection of the parabola and the circle.
For example, for the parabola $y^2 = 3x$ and the circle $x^2 + y^2 = 4$, I can substitute $y^2 = 3x$ into the circle equation and get $x^2 + 3x - 4 = 0$.
This is a quadratic equation that I can factor as $(x+4)(x-1) = 0$, so $x = -4$ or $x = 1$.
Then, using the parabola equation, I can find the corresponding values of $y$.
If $x = -4$, then $y^2 = 3(-4)$, so $y = \pm \sqrt{-12}$, which is not a real number.
So, $x = -4$ does not give a valid solution for the intersection.
If $x = 1$, then $y^2 = 3(1)$, so $y = \pm \sqrt{3}$.
So, $x = 1$ gives two valid solutions for the intersection, namely $(1, \sqrt{3})$ and $(1, -\sqrt{3})$.
These are two of the vertices of the quadrilateral.
Similarly, for the parabola $y^2 = -3x$ and the circle $x^2 + y^2 = 4$, I can substitute $y^2 = -3x$ into the circle equation and get $x^2 - 3x - 4 = 0$.
This is another quadratic equation that I can factor as $(x-4)(x+1) = 0$, so $x = 4$ or $x = -1$.
Then, using the parabola equation, I can find the corresponding values of $y$.
If $x = 4$, then $y^2 = -3(4)$, so $y = \pm \sqrt{-12}$, which is not a real number.
So, $x = 4$ does not give a valid solution for the intersection.
[* a] If $x = -1$, then $y^2 = -3(-1)$, so $y = \pm \sqrt{3}$.
So, $x = -1$ gives two valid solutions for the intersection, namely $(-1, \sqrt{3})$ and $(-1, -\sqrt{3})$.
These are the other two vertices of the quadrilateral.
So, the quadrilateral has vertices $(1, \sqrt{3}), (1, -\sqrt{3}), (-1, \sqrt{3}), (-1, -\sqrt{3})$.
I can sketch these points on a coordinate plane and see that the quadrilateral is actually a rectangle.
To find the area of a rectangle, I need to multiply the length and the width.
The length of the rectangle is the distance between $(1, \sqrt{3})$ and $(-1, \sqrt{3})$, which is $2$ units.
The width of the rectangle is the distance between $(1, \sqrt{3})$ and $(1, -\sqrt{3})$, which is $2\sqrt{3}$ units.