Question

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

• A. $\sqrt{(1-x^2)(1-y^2)}$
• B. $\sqrt{\frac{1-y^2}{1-x^2}}$
• C. $\sqrt{\frac{1-x^2}{1-y^2}}$
• D. None of these

Answer 1

The correct option is B $\sqrt{\frac{1-y^2}{1-x^2}}$

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

Differentiating both sides w.r.t x

$\frac{-x}{\sqrt{1-x^2}}-\frac{y}{\sqrt{1-y^2}}\frac{dy}{dx}=a(1-\frac{dy}{dx})...\left(ii\right)$

put value of a from (I) in(II)

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

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

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

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

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

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

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