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(B) Let $x = \sin 	heta$ and $y = \sin \alpha$. Substituting these into the given equation $y \sqrt{1-x^{2}} = k - x \sqrt{1-y^{2}}$,we get: $\sin \a

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$-\frac{\sqrt{5}}{2}$

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(B) Let $x = \sin 	heta$ and $y = \sin \alpha$. Substituting these into the given equation $y \sqrt{1-x^{2}} = k - x \sqrt{1-y^{2}}$,we get: $\sin \alpha \cos 	heta = k - \sin 	heta \cos \alpha$ $\sin \alpha \cos 	heta + \cos \alpha \sin 	heta = k$ $\sin(\alpha + 	heta) = k$ $\alpha + 	heta = \sin^{-1} k$ Substituting back,we have $\sin^{-1} y + \sin^{-1} x = \sin^{-1} k$. Differentiating both sides with respect to $x$: $\frac{1}{\sqrt{1-x^{2}}} + \frac{1}{\sqrt{1-y^{2}}} \frac{dy}{dx} = 0$ At $x = \frac{1}{2}$,$y = -\frac{1}{4}$. $\sqrt{1-x^{2}} = \sqrt{1 - (\frac{1}{2})^{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$. $\sqrt{1-y^{2}} = \sqrt{1 - (-\frac{1}{4})^{2}} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$. Substituting these values: $\frac{1}{\sqrt{3}/2} + \frac{1}{\sqrt{15}/4} \frac{dy}{dx} = 0$ $\frac{2}{\sqrt{3}} + \frac{4}{\sqrt{15}} \frac{dy}{dx} = 0$ $\frac{4}{\sqrt{15}} \frac{dy}{dx} = -\frac{2}{\sqrt{3}}$ $\frac{dy}{dx} = -\frac{2}{\sqrt{3}} 	imes \frac{\sqrt{15}}{4} = -\frac{1}{2} 	imes \sqrt{5} = -\frac{\sqrt{5}}{2}$.

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

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