If (√(x^2 −1)) +(√(y^2 −1)) =(1/2)(x+y) show that (dy/dx)+(√((y^2 −1)/(x^2 −1))) =0


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Question Number 53676 by peter frank last updated on 24/Jan/19

$I f \sqrt{x^{2} - 1} + \sqrt{y^{2} - 1} = \frac{1}{2} (x + y)$ $s h o w t h a t$ $\frac{d y}{d x} + \sqrt{\frac{y^{2} - 1}{x^{2} - 1}} = 0$
Commented by math1967 last updated on 25/Jan/19

$I t h i n k i t s h o u d b e \frac{d y}{d x} + \frac{y}{x} \sqrt{\frac{y^{2} - 1}{x^{2} - 1}} = 0$
Commented by peter frank last updated on 25/Jan/19

$c a n y o u s h o w i t s i r$
Commented by math1967 last updated on 25/Jan/19

$i t i s n o t t r u e f o r \frac{d y}{d x} + \frac{y}{x} \sqrt{\frac{y^{2} - 1}{x^{2} - 1}} = 0$ $i t i s m y m i s t a k e$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

$l e t f i n d t r u e q u e s t i n . . .$ $1) \frac{d y}{d x} + \sqrt{\frac{y^{2} - 1}{x^{2} - 1}} = 0$ $\frac{d y}{\sqrt{y^{2} - 1}} + \frac{d x}{\sqrt{x^{2} - 1}} = 0$ $\int \frac{d y}{\sqrt{y^{2} - 1}} + \int \frac{d x}{\sqrt{x^{2} - 1}} = c$ $l n (y + \sqrt{y^{2} - 1}) + l n (x + \sqrt{x^{2} - 1)} = l n c$ $(y + \sqrt{y^{2} - 1}) (x + \sqrt{x^{2} - 1}) = c$ $2) \frac{d y}{d x} + \frac{y \sqrt{y^{2} - 1}}{x \sqrt{x^{2} - 1}} = 0$ $\frac{d y}{y \sqrt{y^{2} - 1}} + \frac{d x}{x \sqrt{x^{2} - 1}} = 0$ $\int \frac{d y}{y \sqrt{y^{2} - 1}} + \int \frac{d x}{x \sqrt{x^{2} - 1}} = c$ $y = s e c θ_{1}$ $\int \frac{s e c θ_{1} t a n θ_{1} d θ_{1}}{s e c θ_{1} t a n θ_{1}} + \int \frac{s e c θ_{2} t a n θ_{2} d θ_{2}}{s e c θ_{2} t a n θ_{2}} = c$ $θ_{1} + θ_{2} = c$ ${s e c}^{- 1} (y) + {s e c}^{- 1} (x) = c$ ${c o s}^{- 1} (\frac{1}{y}) + {c o s}^{- 1} (\frac{1}{x}) = c$ $c o s [{c o s}^{- 1} (\frac{1}{y}) + {c o s}^{- 1} (\frac{1}{x})] = c o s c$ $\frac{1}{y} \times \frac{1}{x} - \sqrt{1 - \frac{1}{y^{2}}} \times \sqrt{1 - \frac{1}{x^{2}}} = c o s c$ $\frac{1}{x y} - \frac{\sqrt{y^{2} - 1} \times \sqrt{x^{2} - 1}}{x y} = c o s c$ $1 - \sqrt{y^{2} - 1} \times \sqrt{x^{2} - 1} = x y \times c_{1}$ $s o p l s r e c h e c k y o u r a u e s t i o n$
Commented by peter frank last updated on 25/Jan/19

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