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 $\frac{dy}{dx}$ 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

Answer: (A)

Put $x = sin\theta, y = sin\alpha$
$y\sqrt{1-x^{2}}=k-x\sqrt{1-y^{2}}$
$\Rightarrow sin\alpha \cdot cos\theta+cos\alpha \cdot sin\theta=k$
$\Rightarrow sin\left(\alpha+\theta\right)=k$
$\Rightarrow \alpha+\theta=sin^{-1}\,k$
$\Rightarrow sin^{-1}\,x+ sin^{-1}\,y= sin^{-1}\,k$
$\Rightarrow \frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-y^{2}}}\times\frac{dy}{dx}=0$
at $x=\frac{1}{2}, y=\frac{-1}{4}$
$\frac{dy}{dx}=\frac{-\sqrt{5}}{2}$