Given, \(x=\sin \theta\) and \(y=\cos p \theta\)
\(\therefore \quad \frac{d y}{d \theta}=-p \sin p \theta\) and \(\frac{d x}{d \theta}=\cos \theta\)
\(\Rightarrow \quad \frac{d y}{d x}=-\frac{p \sin p \theta}{\cos \theta}=-p \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2} \frac{d y}{d x}=-p \sqrt{1-y^2}\)
\(\Rightarrow \quad\left(\mathrm{l}-x^2\right)\left(\frac{d y}{d x}\right)^2=p^2\left(\mathrm{l}-y^2\right)\)
(On squaring both sides)
\(\Rightarrow-2 x\left(\frac{d y}{d x}\right)^2+\left(\mathrm{l}-x^2\right) 2 \frac{d y}{d x}\left(\frac{d^2 y}{d x^2}\right)=-p^2 2 y \frac{d y}{d x}\)
(On differentiating both sides)
\(\Rightarrow \quad-x \frac{d y}{d x}+\left(1-x^2\right) \frac{d^2 y}{d x^2}=-p^2 y\)
\(\Rightarrow \quad\left(\mathrm{l}-x^2\right) y_2=x y_1-p^2 y\)
Hence, option (a) is correct.