The equation of tangent to the curve given by $x = a \sin^3 t, y=b \cos ^3 t$ at a point where $t=\frac{\pi}{2}$ is :

y = 0

$x = a \sin^3 t$

$y=b \cos ^3 t$

differentiating both wrt t

so  $\frac{d y}{d t}=-3 b \cos ^2 t \sin t$        .....(1)

$\frac{d x}{d t}=3 a \sin ^2 t \cos t$        .....(2)

so dividing eq (1) by (2)

we get $\frac{d y}{d x}=\frac{-3 b \cos ^2 t \sin t}{3 a \sin^2 t \cos t}=\frac{-b}{a} \frac{\cos t}{\sin t}$

So  $\left.\frac{d y}{d x}\right]_{t=\frac{\pi}{2}}=\frac{-b ~\times 0}{a ~\times 0} =0$ (Slope)

at $t = \frac{\pi}{2}$

y = 0       

x = a

equation of tangent → (x - x₀) × slope = y - y₀

So (x - a) × 0 = y - 0

So y = 0