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Show that the product of the distances from the foci to the tangent at any point of an ellipse is $b^2$ where $b$ is the semi-minor axis.

My process of proof so far:

Let $F(c, 0)$ and $F'(-c, 0)$ be 2 points that lie on the x-axis of the plane and $P(x_0, y_0)$ be the point that lies on the tangent to the ellipse. I simply started by simplifying the equation $$\sqrt{(x+c)^2+y^2}\sqrt{(x-c)^2+y^2}$$

However, this got me to a dead end. Plus, I feel as if there's some property about the tangent of an ellipse that I'm not using. If anyone could guide (not answer) me towards the answer, that would be extremely beneficial to my sanity (and knowledge, of course).

Thanks again.

S. Sharma
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  • As stated, the statement is (obviously) false: "any point of any tangent" can be arbitrarily-far from the ellipse, so the product of distances to such points is non-constant (and is, indeed, unbounded). You probably mean to write "the product of the distances from the foci to the tangent at any point of an ellipse is $b^2$ [...]". (I suppose you could mean something else, but at least this statement happens to be true.) – Blue Nov 06 '18 at 05:23
  • By the way: If the proper statement is as I suggested in my previous comment, then this question is a duplicate of "Ellipse Focal Proof". – Blue Nov 06 '18 at 07:07
  • That's what I think the question was trying to ask. I got this problem out of a book, so I was quite confused when the answer wasn't $b^2$. – S. Sharma Nov 06 '18 at 12:27
  • Okay. Then I'm going to vote to close this question as a duplicate of the other one. – Blue Nov 06 '18 at 12:31

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Hint

WLOG the equation of the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

where $b^2=a^2(1-e^2)$

The coordinates of the foci are $(\pm ae,0)$

What is the equation of tangent at $P(a\cos t,b\sin t)$

What are the perpendicular distances of the foci from the tangent?

lab bhattacharjee
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