Question

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

• A. $-\sqrt{\frac{1-y^2}{1-x^2}}$
• 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 A $-\sqrt{\frac{1-y^2}{1-x^2}}$

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

By differentiating , we get

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

On solving this, we get

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