Question

If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$

Answer 1

Given

$\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$

$ay+\sqrt{1-y^2}=ax-\sqrt{1-x^2}$

Differentiate on both sides w.r.t. x

$a\frac{dy}{dx}+\frac{-2y}{2\sqrt{1-y^2}}\frac{dy}{dx}=a-\frac{-2x}{2\sqrt{1-x^2}}$

$\frac{dy}{dx}\left[\frac{a\sqrt{1-y^2}-y}{a\sqrt{1-x^2}+x}\right]=\sqrt{\frac{1-y^2}{1-x^2}}$

$a\sqrt{1-y^2}-y=\frac{\sqrt{(1-x^2)(1-y^2)}+1-y^2-xy+y^2}{(x-y)}$

$=\frac{\sqrt{(1-x^2)(1-y^2)}+1-xy+x^2-x^2}{(x-y)}$

$=\frac{\sqrt{(1-x^2)(1-y^2)}+(1-x^2)+x(x-y)}{(x-y)}$

$=\sqrt{1-x^2}\frac{(\sqrt{1-y^2}+\sqrt{1-x^2})}{(x-y)}+x$

$\therefore a\sqrt{1-y^2}-y=a\sqrt{1-x^2}+x$

$\therefore \frac{dy}{dx}\left(1\right)=\sqrt{\frac{1-y^2}{1-x^2}}$

$\therefore \frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$