If $x = a \cos t$ and $y = b \sin t$, then find $\frac{d^2y}{dx^2}$.

$-\frac{b}{a^2} \text{cosec}^3 t$

The correct answer is Option (2) → $-\frac{b}{a^2} \text{cosec}^3 t$ ##

Given, $x = a \cos t, y = b \sin t$

$\frac{dx}{dt} = -a \sin t$

$ \frac{dy}{dt} = b \cos t$

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b \cos t}{-a \sin t}$

$= -\frac{b}{a} \cot t$

$\frac{d^2y}{dx^2} = -\frac{b}{a} \frac{d}{dx} \cot t$

$= +\frac{b}{a} (-\text{cosec}^2 t) \frac{dt}{dx}$

$= \frac{b}{a} \text{cosec}^2 t \times \frac{1}{-a \sin t}$

$= -\frac{b}{a^2} \text{cosec}^3 t$