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

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

Differentiating both sides wrt x,

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

or, $\frac{-x}{\sqrt{1-x^2}}-\frac{y}{\sqrt{1-y^2}}\frac{dy}{dx}=a(1-\frac{dy}{dx})$

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

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

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

(from $\left(1\right)$)

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

Proved.