🛠️ JEE➗ Maths

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 const…

Q1

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 \( \frac{\mathrm{dy}}{\mathrm{dx}} \) at \( x=\frac{1}{2} \), is equal to :

[JEE Main 2020, 7 Jan (Shift 2)]

a

\( \frac{\sqrt{5}}{2} \)

b

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

c

\( \frac{2}{\sqrt{5}} \)

d

\( \frac{-\sqrt{5}}{4} \)

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