Let $y=y(x)$ 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 $\frac{d y}{d x}$ at $x=\frac{1}{2}$, is equal to:
Correct Option: , 2
Given, $x=\frac{1}{2}, y=\frac{-1}{4} \Rightarrow x y=\frac{-1}{8}$
$y \cdot \frac{1 \cdot(-2 x)}{2 \sqrt{1-x^{2}}}+y^{\prime} \sqrt{1-x^{2}}$
$=-\left\{1 \cdot \sqrt{1-y^{2}}+\frac{x \cdot(-2 y)}{2 \sqrt{1-y^{2}}} y^{\prime}\right\}$
$\Rightarrow \quad-\frac{x y}{\sqrt{1-x^{2}}}+y^{\prime} \sqrt{1-x^{2}}=-\sqrt{1-y^{2}}+\frac{x y \cdot y^{\prime}}{\sqrt{1-y^{2}}}$
$\Rightarrow y^{\prime}\left(\sqrt{1-x^{2}}-\frac{x y}{\sqrt{1-y^{2}}}\right)=\frac{x y}{\sqrt{1-x^{2}}}-\sqrt{1-y^{2}}$
$\Rightarrow y^{\prime}\left(\frac{\sqrt{3}}{2}+\frac{1}{8 \cdot \frac{\sqrt{15}}{4}}\right)=\frac{-1}{8 \cdot \sqrt{\frac{3}{2}}}-\frac{\sqrt{15}}{4}$
$\Rightarrow y^{\prime}\left(\frac{\sqrt{45}+1}{2 \sqrt{15}}\right)=-\frac{(1+\sqrt{45})}{4 \sqrt{3}}$
$\therefore \quad y^{\prime}=-\frac{\sqrt{5}}{2}$