Limits, Continuity & Differentiability Question - None

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

Accepted Answer

Step 1: Given functional relation is $y\sqrt{1-x^2} = k - x\sqrt{1-y^2}$

Step 2: On differentiating both sides w.r.t. x, we get:

$\frac{d}{dx}\left(y\sqrt{1-x^2}\right) = \frac{d}{dx}\left(k - x\sqrt{1-y^2}\right)$

Step 3: Using product rule and chain rule:

$\sqrt{1-x^2}\frac{dy}{dx} + y\cdot\frac{-2x}{2\sqrt{1-x^2}} = 0 - \sqrt{1-y^2} - x\cdot\frac{-2y}{2\sqrt{1-y^2}}\cdot\frac{dy}{dx}$

Step 4: Simplifying:

$\sqrt{1-x^2}\frac{dy}{dx} - \frac{xy}{\sqrt{1-x^2}} = -\sqrt{1-y^2} + \frac{xy}{\sqrt{1-y^2}}\frac{dy}{dx}$

Step 5: Rearranging:

$\left(\sqrt{1-x^2} - \frac{xy}{\sqrt{1-y^2}}\right)\frac{dy}{dx} = -\sqrt{1-y^2} + \frac{xy}{\sqrt{1-x^2}}$

Step 6: At $x = \frac{1}{2}$, $y = -\frac{1}{4}$:

$\sqrt{1-\frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

$\sqrt{1-\frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$

Step 7: Substituting and solving yields $\frac{dy}{dx} = -\frac{5}{2}$

∴ Answer is (b).

Question:

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