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Tangent at any point $P$ on an ellipse whose foci are $F_1,F_2$ meets the auxiliary circle of the ellipse at $B_1$, $B_2$. If $F_{1}P+F_{2}P=10$ and $(F_{1}B_{1}) \cdot(F_{2}B_{2})=16$, then eccentricity of the ellipse is equal to?

In this question I understand that $2a=10$ considering the form as $\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$.What does $(F_{1}B_{1}) \cdot (F_{2}B_{2})=16$ imply?

Please give your suggestions !

balddraz
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2 Answers2

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What does $(F_{1}B_{1}) \cdot (F_{2}B_{2})=16$ imply ? Please give your suggestions !

Hint: Let P coincide with the “tip” of the minor semi-axis. What do you notice ? :-$)$

Lucian
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  • That's only a special case ! Does that always hold true?Why? –  Oct 12 '15 at 06:29
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    @SanchayanDutta: Because the product is constant. So, since its value is always the same, no matter where P is on the ellipse, might as well choose a position that's helpful. – Lucian Oct 12 '15 at 06:32
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$ a = 5, b= 4,$ by the property of perpendicular segments products constancy $ = b^2$

$$ \epsilon = \sqrt{ 1- 16/25} = 3/5 $$

Property constancy of focal ray projections

$$ r_1 r_2 \sin ^2 \psi = b^2 $$

can be easily proved, where $ r_1,r_2 $ are distances to each focus and $\psi$ is tangent-focal ray (mirror reflection) angles at each point.

Narasimham
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