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Question Number 39955 by rahul 19 last updated on 13/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18

	$f o r h y p e r b o l a \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1 B^{2} = A^{2} (e_{2}^{2} - 1)$ $f o r e l l i p s e b^{2} = a^{2} (1 - e_{1}^{2})$ $f o c i o f h y e r b o l a = ({A e}_{2}, 0) a n d (- {A e}_{2}, 0)$ $v e r t i c e s o f e l l i p s e = (a, 0) a n d (- a, 0)$ $f o c i o f e l l i p s e ({a e}_{1}, 0) a n d (- {a e}_{1}, 0)$ $g i v e n {a e}_{1} = A s o \frac{a}{A} = \frac{1}{e_{1}}$ ${A e}_{2} = a \frac{a}{A} = e_{2}$ $\frac{1}{e_{1}} = e_{2}$ ${(\frac{b}{B})}^{2} = \frac{a^{2} (1 - e_{1}^{2})}{A^{2} (e_{2}^{2} - 1)} = \frac{e_{2}^{2} (1 - \frac{1}{e_{2}^{2}})}{(e_{2}^{2} - 1)} = 1$
	Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18

	$e_{1} + e_{2} = {(\sqrt{e_{1}} - \sqrt{e_{2}})}^{2} + 2 \sqrt{e_{1} e_{2}}$ $= {(\sqrt{e_{1}} - \sqrt{e_{2}})}^{2} + 2 (s i n c e e_{1} e_{2} = 1 a l r e a d y p r o v e d)$ $e_{1} + e_{2} > 2 s o e_{1} + e_{2} c a n n o t b e e q u a l s t o 2$
	Commented by rahul 19 last updated on 15/Jul/18
	Thank You Sir ! ��
	Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

	$i t s o k . . . a t t h e a g e o f 48 s t i l l i n t e r e s t e d i n$ $p h y s i c s a n d m a t h . . . l o o k b a c k . . . t h e y e a r 1988 .$ $w h e n i w a s i n c l a s s X I I . .$
	Commented by rahul 19 last updated on 15/Jul/18
	That's the spirit! ��
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