Question

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

Answer 1

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

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

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

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

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

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

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