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Question Number 45608 by peter frank last updated on 14/Oct/18

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

	$p (α, β) Q (a, b) p o i n t R d e v i d e s P Q 2 : 1$ $R (\frac{2 a + α}{3}, \frac{2 b + β}{3}) = (h, k) s a y$ $\frac{2 a + α}{3} = h α = 3 h - 2 a$ $\frac{2 b + β}{3} = k β = 3 k - 2 b$ $s i n c e α, β l i e s o n h y p e r b o l a \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ $s o \frac{α^{2}}{a^{2}} - \frac{β^{2}}{b^{2}} = 1$ $c a u t i o n h e r e c o - o r d i n a t e o f p o i n t Q i s g i v e n$ $(a, b) w h e r a s i n h y p e r b o l a e q n a l r e a d y a, b$ $e x i s t s . . . s o p l s c h e c k w h e t h e r t h e y a r e i d e n t i c a l$ $. . .$ $\frac{{(3 h - 2 a)}^{2}}{a^{2}} - \frac{{(3 k - 2 b)}^{2}}{b^{2}} = 1$ $s o t h e r e q u i r e d l o c u s i s$ $\frac{{(3 x - 2 a)}^{2}}{a^{2}} - \frac{{(3 y - 2 b)}^{2}}{b^{2}} = 1 p l s c h e c k . . .$
	Commented by peter frank last updated on 14/Oct/18

	$thank sir your absolute right$
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