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 $\mathrm{y}\left(\frac{1}{2}\right)=-\frac{1}{4}$. Then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=\frac{1}{2}$, is equal to:

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Derivatives XI

Let

y = y (x)

be a function of

x

satisfying

y 1 - x^{2} = k - x 1 - y^{2}

where

k

is a constant and

y (\frac{1}{2}) = - \frac{1}{4}

. Then

\frac{dy}{dx}

x = \frac{1}{2}

, is equal to:

Answer: (B)

Differentiating

y 1 - x^{2} = k - x 1 - y^{2}

implicitly gives

1 - x^{2} y^{'} - \frac{x y}{1 - x ^{2}} = - 1 - y^{2} + \frac{x y y ^{'}}{1 - y ^{2}}

. At

x = \frac{1}{2}, y = - \frac{1}{4}

the radicals are

1 - x^{2} = \frac{3}{2}

and

1 - y^{2} = \frac{15}{4}

. Substituting these and gathering the

y^{'}

terms leaves a single linear equation for

y^{'}

, which solves to

\frac{d y}{d x} = - \frac{5}{2}

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option