Question

Let $y=y\left(x\right)$ be a function of $x$ satisfying $y\sqrt{1-x^2}=k-x\sqrt{1-y^2}$ where $k$ is a constant and $y\left(\frac{1}{2}\right)=-\frac{1}{4}$. Then $\frac{dy}{dx}$ at $x=\frac{1}{2}$, is equal to :

• A. $-\frac{\sqrt{5}}{2}$
• B. $\frac{\sqrt{5}}{2}$
• C. $-\frac{\sqrt{5}}{4}$
• D. $\frac{2}{\sqrt{5}}$

Answer 1

The correct option is A $-\frac{\sqrt{5}}{2}$

$y\sqrt{1-x^2}=k-x\sqrt{1-y^2}$
Differentiating w.r.t. $x$ on both the sides, we get
$y^{\prime }\sqrt{1-x^2}+y\times \frac{1}{2\sqrt{1-x^2}}\times (-2x)$

$=-\sqrt{1-y^2}-x\times \frac{1}{2\sqrt{1-y^2}}\times (-2y)y^{\prime }$

$\Rightarrow y^{\prime }\sqrt{1-x^2}-\frac{xy}{\sqrt{1-x^2}}=\frac{xy}{\sqrt{1-y^2}}{y}^{\prime }-\sqrt{1-y^2}$

Putting $x=\frac{1}{2},y=-\frac{1}{4}$, we get
$y^{\prime }[\frac{\sqrt{3}}{2}+\frac{\frac{1}{8}}{\frac{\sqrt{15}}{4}}]=-\frac{\frac{1}{8}}{\frac{\sqrt{3}}{2}}-\frac{\sqrt{15}}{4}$

$\Rightarrow y^{\prime }[\frac{\sqrt{3}}{2}+\frac{1}{2\sqrt{15}}]=-\frac{1}{4\sqrt{3}}-\frac{\sqrt{15}}{4}$

$\Rightarrow y^{\prime }\left[\frac{\sqrt{45}+1}{2\sqrt{15}}\right]=-\frac{1+\sqrt{45}}{4\sqrt{3}}$
$\Rightarrow y^{\prime }{|}_{x=1/2}=-\frac{\sqrt{5}}{2}$