Explanation

(d)
Given, $x = \sin t,\ \ y = \cos{pt}$
$\frac{dx}{dt} = \cos t,\ \ \frac{dy}{dt} = - p\sin{pt}$
$\therefore$ $\frac{dy}{dx} = - \frac{p\sin{pt}}{\cos t}$
$\Rightarrow$ $y_{1} = \frac{- p\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}}$
$\Rightarrow$ $y_{1}\sqrt{1 - x^{2}} = - p\sqrt{1 - y^{2}}$
$\Rightarrow$ $y_{1}^{2}\left( 1{- x}^{2} \right) = p^{2}(1 - y^{2})$
$\Rightarrow$ $2y_{1}y_{2}\left( 1 - x^{2} \right) - 2xy_{1}^{2} = - 2yy_{1}p^{2}$ [differentiating]
$\Rightarrow$ $\left( 1 - x^{2} \right)y_{2} - xy_{1} + p^{2}y = 0$