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