Question

If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$,then $\frac{dy}{dx}$ is equal to

Answer 1

rearranging we have

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

let $x=\sin p$ and $y=\sin q$

$\therefore \frac{\cos p+\cos q}{\sin p-\sin q}=a$

$\frac{2\cos \frac{p+q}{2}\cos \frac{p-q}{2}}{2\cos \frac{p+q}{2}\sin \frac{p-q}{2}}=a$

$\cot \frac{p-q}{2}=a$

$p-q=2{\cot }^{-1}a$

${\sin }^{-1}x-{\sin }^{-1}y=2{\cot }^{-1}a$

differentiating the above equation wrt $x$ we have

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

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