Show that the locus of a point which moves so that its distance from the point (ae,0) is e times its distance from the line x=(a/e) is given by the equation (x^2 /a^2 )+(y^2 /(a^2 (1−e^2 )))=1


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Question Number 51271 by peter frank last updated on 25/Dec/18

$S h o w t h a t t h e l o c u s o f a$ $p o i n t w h i c h m o v e s s o$ $t h a t i t s d i s t a n c e f r o m$ $t h e p o i n t (a e, 0) i s e t i m e s$ $i t s d i s t a n c e f r o m t h e$ $l i n e x = \frac{a}{e} i s g i v e n b y t h e$ $e q u a t i o n$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 - e^{2})} = 1$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18

	$\sqrt{{(x - a e)}^{2} + y^{2}} = e (x - \frac{a}{e})$ $\sqrt{{(x - a e)}^{2} + y^{2}} = (e x - a)$ $x^{2} - 2 x a e + a^{2} e^{2} + y^{2} = e^{2} x^{2} - 2 x a e + a^{2}$ $x^{2} (1 - e^{2}) + y^{2} = a^{2} (1 - e^{2})$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 - e^{2})} = 1$
	Commented by peter frank last updated on 25/Dec/18

	$m u c h r e s p e c t m r t a n m a y .$ $G O D b l e s s y o u .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18

	$t h a n k y o u . . .$
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