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Question Number 45602 by peter frank last updated on 14/Oct/18

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

	$(c t, \frac{c}{t}) l i e s o n r e c t a n g u l s r h y p e r b o l a$ $x y = c^{2}$ $\frac{d}{d x} (x y) = \frac{d}{d x} (c^{2})$ $x \frac{d y}{d x} + y = 0 \frac{d y}{d x} = - \frac{y}{x}$ $s l o p e o f t s n g e n t a t (c t, \frac{c}{t})$ $m = {(\frac{d y}{d x})}_{(c t, \frac{c}{t})} = \frac{- \frac{c}{t}}{c t} = - \frac{1}{t^{2}}$ $m \times m^{'} = - 1$ $- \frac{1}{t^{2}} \times m^{'} = - 1 m^{'} = t^{2}$ $n o r m a l e q n$ $y - \frac{c}{t} = t^{2} (x - c t)$ $p l s w a i t . . .$
	Commented by peter frank last updated on 17/Oct/18

	$am still waiting sir . . . .$
	Answered by ajfour last updated on 14/Oct/18

	$x y = c^{2}$ $A p o i n t o n t h e h y p e r b o l a (c t, \frac{c}{t})$ $- \frac{d x}{d y} = t^{2}$ $l e t p o i n t f r o m w h e r e t h e n o r m a l s$ $a r e d r a w n b e (h, k)$ $e q . o f n o r m a l s :$ $y - k = t^{2} (x - h)$ $s i n c e (c t, \frac{c}{t}) l i e s o n s u c h n o r m a l,$ $s o \frac{c}{t} - k = t^{2} (c t - h)$ $\Rightarrow {c t}^{4} - {h t}^{3} + k t - c = 0$ $. . . . . . .$
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