Question

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

Answer 1

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

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

Differentiating both sides with respect to x, we get

$\Rightarrow \frac{(x-y)\left\{\frac{1}{2\sqrt{1-x^2}}\right(2-x)+\frac{(-2y)}{2\sqrt{1-y^2}}\frac{dy}{dx}\}-(\sqrt{1-x^2}+\sqrt{1-y^2})(1-\frac{dy}{dx})}{(x-y{)}^2}=0$

$\Rightarrow (x-y)(\frac{-x}{\sqrt{1-x^2}}-\frac{y}{\sqrt{1-y^2}}\frac{dy}{dx})-(\sqrt{1-x^2}+\sqrt{1-y^2})(1-\frac{dy}{dx})=0$

$\Rightarrow \frac{-x(x-y)}{\sqrt{1-x^2}}-\frac{(x-y)y}{\sqrt{1-y^2}}\frac{dy}{dx}-(\sqrt{1-x^2}+\sqrt{1-y^2})+(\sqrt{1-x^2}+\sqrt{1-y^2})\frac{dy}{dx}=0$

$\Rightarrow \frac{-x(x-y)}{\sqrt{1-x^2}}-\sqrt{1-x^2}-\sqrt{1-y^2}+(\sqrt{1-x^2}+\sqrt{1-y^2}-\frac{(x-y)y}{\sqrt{1-y^2}})\frac{dy}{dx}=0$

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

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

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

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

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