Problem Statement:
Square $ABCD$ has side length $60$. An ellipse $E$ is circumscribed about the square and there is a point $P$ on the ellipse such that $PC = PD =50$. What is the area of $E$?
I have absolutely no idea nor any thoughts I can write down here. How would I approach this problem? How to do it?
Hint
Let's write the ellipse like:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
And the vertex: $A(-30,30),B(30,30),C(30,-30),D(-30,-30),$ so once those belong to the ellipse then
$$\frac{900}{a^2}+\frac{900}{b^2}=1$$
We also have, $PC=50=PD$, $P(x_0,y_0)$:
$$(x_0-30)^2+(y_0+30)^2=50^2\\ (x_0+30)^2+(y_0+30)^2=50^2$$
After find $P(x_0,y_0)$ and remembering that it belongs to the elipse you can find $a$ and $b$.
And the area will be given by $\pi \cdot a\cdot b$.
Can you finish?