One way is to find the foot of perpendicular and directly putting it into the equation of auxiliary circle. But that is quite a lengthy proof, is there any other short method to prove this property?
2026-04-13 17:40:52.1776102052
How to prove that the foot of perpendicular drawn from a focus to any tangent of an ellipse lies on its auxiliary circle?
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OUTLINE
Let the foci be $F,F'$. Let $P$ be a point on the ellipse and let the feet for the perpendiculars from $F,F'$ to the tangent at $P$ be $T,T'$. Let the lines $F'P,FT$ meet at $X$. Let $O$ be the centre of the ellipse.
Show that $FPT,XPT$ are congruent. Using the focal distances property show that $F'X=2a$. Prove that $OT$ is parallel to $F'X$. Prove that $OT=a$.