Area of intersection of two ellipses

Here's a messy, brute-force solution. The problem is flagged as geometry, but I definitely resorted to calculus, which makes me think there's a better way to do this!

The general equation for a conic section in Cartesian coordinates is $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ . For an ellipse with semi-major axis $a$ , semi-minor axis $b$ , centre $(x_c,y_c)$ and with its major axis an angle $\theta$ from the positive $x$ -axis, these coefficients are

$\begin{aligned} A &= a^2 \sin^2 \theta + b^2 \cos^2 \theta \\ B &= 2(b^2-a^2)\sin \theta \cos \theta \\ C &= a^2 \cos^2 \theta + b^2 \sin^2 \theta \\ D &= -2Ax_c - By_c \\ E &= -Bx_c - 2Cy_c \\ F &= Ax_c^2 + Bx_c y_c + Cy_c^2 - a^2 b^2 \end{aligned}$

(these can be derived by applying coordinate transforms to an ellipse centred at $O$ with its axes on the coordinate axes, or found here )

Using the given information, for the blue ellipse, we have

$\begin{aligned} x_c &= y_c = 0 \\ a^2 &= 10^2+(-30)^2=1000 \\ b^2 &= (a/2)^2=250 \\ \cos \theta &= \frac{-10}{\sqrt{10^2+(-30)^2}} = \frac{-1}{\sqrt{10}} \\ \sin \theta &= \frac{30}{\sqrt{10^2+(-30)^2}} = \frac{3}{\sqrt{10}} \end{aligned}$

Substituting in and cancelling common factors, we get the equation $37x^2 + 18x y + 13y^2 - 10000 = 0$ for the blue ellipse.

In the same way, we find $3796x^2 - 4480x y + 2164y^2 + 63780x - 54480y - 447111 = 0$ for the orange ellipse.

These implicit equations can be thought of as quadratics in the variable $y$ ; for example, the equation for the blue ellipse can be solved for $y$ to give

$\begin{aligned} y &= \frac{\pm 20 \sqrt{325 - x^2} - 9 x}{13} \end{aligned}$

For simplicity, let's call these two solutions $y=B_+(x)$ and $y=B_-(x)$ and define $y=O_+(x)$ and $y=O_-(x)$ for the orange ellipse in the same way.

Here I started using numerical methods. The equations can be solved to find the four points where the ellipses intersect; these are $(-17.86,8.54)$ , $(-10.08,-16.02)$ , $(-2.51,29.20)$ and $(12.63,11.06)$ . Call these $X_i,Y_i$ for $i=1,2,3,4$ . Using these, the pink area can be divided into three distinct regions, as below:

The red region is bounded by $y=O_+$ above and $y=B_-$ below, so its area is

$\begin{aligned} \int_{X_1}^{X_2} (O_+-B_-) dx \approx 167.22 \end{aligned}$

Similarly, the green area is

$\begin{aligned} \int_{X_2}^{X_3} (O_+-O_-) dx \approx 285.60 \end{aligned}$

and the yellow area is

$\begin{aligned} \int_{X_3}^{X_4} (B_+-O_-) dx \approx 334.54 \end{aligned}$

These integrals can be computed analytically, but I went with numerical methods. Summing, we get the answer: $167.22+285.60+334.54=\boxed{787.36}$