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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?

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    $\begingroup$ I think you mean: foot of the perpendicular on a tangent line to the ellipse ? This should be added... $\endgroup$ – Jean Marie Apr 28 '16 at 8:46
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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 $OT=a$.

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First prove that the reflection of Focus S1 with respect to any tangent lies on a circle with focus S2 as center and 2a as radius .let the reflection be M Now in triangle S1S2M, see that foot of perpendicular is midpoint of S1M and center is midpoint of S1S2 so by midpoint theorem we can see that CX=1/2(2a) where X is foot of perpendicular so CX=a . So X lies on auxillary circle

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