Question

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

Answer 1

Understanding differentiation using gradient formula for quadratic graphs

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

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

let $x=\sin A;y=\sin B...\left(1\right)$

$f\left(x\right)=\sqrt{1-{\sin }^{2}A}+\sqrt{1-{\sin }^{2}B}=a(\sin A-\sin B)=\cos A+\cos B=a(\sin A-\sin B)$

we know that $[1-{\sin }^{2}A={\cos }^{2}A]$

$\Rightarrow 2\cos \left(\frac{A+B}{2}\right)-\cos \left(\frac{A-B}{2}\right)=a\left[2\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)\right]$

$\Rightarrow 2\cos \left(\frac{A-B}{2}\right)=a.2\sin \left(\frac{A-B}{2}\right)\Rightarrow \frac{\cos \left(\frac{A-B}{2}\right)}{\sin \left(\frac{A-B}{2}\right)}=a$

$\Rightarrow \cot \left(\frac{A-B}{2}\right)=a[\because \frac{\cos \theta }{\sin \theta }=\cot \theta ]$

$\Rightarrow \frac{A-B}{2}={\cot }^{-1}a$

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

Applying differentiation on both sides

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

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