• question_answer 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 [JEE MAIN Held on 07-01-2020 Evening] A) $-\frac{\sqrt{5}}{2}$ B) $\frac{2}{\sqrt{5}}$ C) $\frac{\sqrt{5}}{2}$  D) $-\frac{\sqrt{5}}{4}$

Solution :

[a] $\because \,\,\,\,y\sqrt{1-{{x}^{2}}}=k-x\sqrt{1-{{y}^{2}}}$                ...(i) On differentiating both side of eq. (i) w.r.t. x we get, $\frac{dy}{dx}\sqrt{1-{{x}^{2}}}-y\frac{2x}{2\sqrt{1-{{x}^{2}}}}=0-\sqrt{1-{{y}^{2}}}+\frac{x\cdot y}{\sqrt{1-{{y}^{2}}}}\frac{dy}{dx}$ Put $x=\frac{1}{2}$and $y=-\frac{1}{4}$ we get $\frac{dy}{dx}\cdot \frac{\sqrt{3}}{2}-\left( -\frac{1}{4} \right)\cdot \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{\sqrt{15}}{4}+\frac{-\frac{1}{8}}{\frac{\sqrt{15}}{4}}\cdot \frac{dy}{dx}$ $\therefore \,\,\,\frac{dy}{dx}=-\frac{\sqrt{5}}{2}$

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