(√(1−x^2 )) + (√(1−y^2 )) = a(x−y) prove that (dy/dx)=(√((1−y^2 )/(1−x^2 )))


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Question Number 54586 by arvinddayama00@gmail.com last updated on 07/Feb/19

$\sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y)$ $p r o v e t h a t$ $\frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Feb/19

	$x = s i n p y = s i n q$ $\frac{d x}{d p} = c o s p \frac{d y}{d q} = c o s q$ $\frac{d y}{d x} = \frac{c o s q d q}{c o s p d p}$ $\sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y)$ $c o s p + c o s q = a (s i n p - s i n q)$ $a = \frac{2 c o s \frac{p + q}{2} \times c o s \frac{p - q}{2}}{2 c o s \frac{p + q}{2} s i n \frac{p - q}{2}}$ $t a n \frac{p - q}{2} = \frac{1}{a}$ ${s e c}^{2} (\frac{p - q}{2}) (\frac{\frac{d p}{d x} - \frac{d q}{d x}}{2}) = 0$ $s o \frac{d p}{d x} - \frac{d q}{d x} = 0 s o \frac{d p}{d q} = 1$ $n o w \frac{d y}{d x} = \frac{c o s q}{c o s p} \times \frac{d q}{d p} [\frac{d q}{d p} = 1]$ $\frac{d y}{d x} = \frac{c o s q}{c o s p} = \frac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}}$
	Commented by arvinddayama00@gmail.com last updated on 08/Feb/19

	$a n y o r s o l u t i o n$
	Commented by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19

	$t r y i n g b y o t h e r m e t h o d$
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