Explanation

(c)
Given curves are $x = 2\cos^{3}\theta$ and $y = 3\sin^{3}\theta$
$\therefore\ \ \frac{dx}{d\theta} = - 6\cos^{2}\theta\sin\theta$,
$\frac{dy}{d\theta} = 9\sin^{2}\theta\cos\theta$
$\therefore\ \ \ \ \frac{dy}{dx} = - \frac{9\sin^{2}\theta\cos\theta}{6\cos^{2}\theta\sin\theta} = - \frac{3}{2}\tan\theta$
At $\theta = \frac{\pi}{4},\ \ \ \ \frac{dy}{dx} = - \frac{3}{2}\tan{\frac{\pi}{4} = - \frac{3}{2}}$
Also, at $\theta = \frac{\pi}{4},\ \ x = 2\cos^{3}{\frac{\pi}{4} = - \frac{1}{\sqrt{2}}}$
And $y = 3\sin^{2}\theta = 3\left( \frac{1}{\sqrt{2}} \right)^{3}$
$= \frac{3\sqrt{2}}{4}$
Equation of tangent at $\left( \frac{1}{\sqrt{2}},\frac{3\sqrt{2}}{4} \right)$ is
$\left( y - \frac{3\sqrt{2}}{4} \right) = - \frac{3}{2}\left( x - \frac{1}{\sqrt{2}} \right)$
$\Rightarrow \ \ \ \ 4y - 3\sqrt{2} = - 6x + 3\sqrt{2}$
$\Rightarrow \ \ \ 3x + 2y = 3\sqrt{2}$