Ellipse - relation between a and b such that $F_1P \perp F_2P$

Question

Consider the ellipse $\displaystyle \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}=1$ with foci $F_1 (-e, 0)$ and $F_2 (e, 0)$ (where $e$ is the linear eccentricity).

What is the relation between $a$ and $b$ so that we get points $P$ on the ellipse satisfying $F_1P \perp F_2P$?

I'd be glad for any hint to solve this problem.

Answer 1

There is such a point $P$ on the ellipse iff the system $${x^2\over a^2}+{y^2\over b^2}=1,\qquad x^2+y^2=e^2=a^2-b^2\tag{1}$$ has real solutions $(x,y)$. From $(1)$ one obtains $$x^2={a^2(a^2-2b^2)\over a^2-b^2},\qquad y^2={b^4\over a^2-b^2}\ .$$ This gives real $(x,y)$ iff $a\geq\sqrt{2}\,b$.

Answer 2

HINT :

If $F_1P\perp F_2P$, then the $P$ is on the circle whose diameter is $F_1F_2$.