Test for intersection:
You can stretch space in one axis so that one of the ellipses is transformed in a circle. Then there is intersection if
- the center of the circle is inside the ellipse, or
- the center is outside but the distance between the point and the ellipse outline is smaller than the radius.
Testing insideness is immediate (plug the coordinates of the center in the implicit equation of the ellipse and check the sign).
Computing the distance is another matter. Let the center of the circle be $(x,y)$ and the ellipse be at the origin (WLOG), with semi-axis $a,b$ (sorry for the change of notation, I can't help it). The squared distance between the given point and an arbitrary point on the ellipse outline is
$$d^2=(a\cos(t)-x)^2+(b\sin(t)-y)^2.$$
The extrema of the distance are achieved when the derivative cancels, i.e.
$$-a\sin(t)(a\cos(t)-x)+b\cos(t)(b\sin(t)-y)=0$$
which can be written
$$\frac{ax}{\cos(t)}-\frac{by}{\sin(t)}=a^2-b^2.$$
This equation can be rationalized by the usual transform
$$\cos(t)=\frac{1-z^2}{1+z^2},\sin(t)=\frac{2z}{1+z^2},$$
giving the quartic equation
$$byz^4 + 2(ax + (a^2-b^2))z^3 + 2(ax -(a^2-b^2)^2)z-by= 0.$$
One of the real roots gives the shortest distance.
The $125$ and the $100$ in your hyperbola equation should have been swapped:
$$\frac{x^2}{125}-\frac{y^2}{100}=1$$
This gives $P$ as $\left( \dfrac{25\sqrt5}3,\dfrac{40}{3} \right)$ and $PA \times PB \,$ yields the given answer of $500$.
Best Answer
The pink area can be described by a system of inequations, for instance
$$\begin{cases} \dfrac{(x-x_1)^2}{a_1^2}+\dfrac{y^2}{b_1^2}\le 1,\\ \dfrac{(x-x_2)^2}{a_2^2}+\dfrac{y^2}{b_2^2}\le 1. \end{cases}$$