# Let y=y(x) be a function of x satisfying

Question:

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:

1. (1) $-\frac{\sqrt{5}}{4}$

2. (2) $-\frac{\sqrt{5}}{2}$

3. (3) $\frac{2}{\sqrt{5}}$

4. (4) $\frac{\sqrt{5}}{2}$

Correct Option: , 2

Solution:

Given, $x=\frac{1}{2}, y=\frac{-1}{4} \Rightarrow x y=\frac{-1}{8}$

$y \cdot \frac{1 \cdot(-2 x)}{2 \sqrt{1-x^{2}}}+y^{\prime} \sqrt{1-x^{2}}$

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

$\Rightarrow \quad-\frac{x y}{\sqrt{1-x^{2}}}+y^{\prime} \sqrt{1-x^{2}}=-\sqrt{1-y^{2}}+\frac{x y \cdot y^{\prime}}{\sqrt{1-y^{2}}}$

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

$\Rightarrow y^{\prime}\left(\frac{\sqrt{3}}{2}+\frac{1}{8 \cdot \frac{\sqrt{15}}{4}}\right)=\frac{-1}{8 \cdot \sqrt{\frac{3}{2}}}-\frac{\sqrt{15}}{4}$

$\Rightarrow y^{\prime}\left(\frac{\sqrt{45}+1}{2 \sqrt{15}}\right)=-\frac{(1+\sqrt{45})}{4 \sqrt{3}}$

$\therefore \quad y^{\prime}=-\frac{\sqrt{5}}{2}$