Area of the curve $y=\sin\theta \cos^2\theta$; $x=\sin^2\theta \cos\theta$ bounded by tangent parallel to axes Area bounded by tangents of the curve given by $y=\sin\theta \cos^2\theta$; $x=\sin^2\theta \cos\theta$, which are parallel to co-ordinate axes(other than axes) is 
(1)$\frac{4}{27}$
(2) $\frac{27}{4}$
(3)$\frac{16}{27}$
(4) $\frac{27}{16}$
My approach is as follow $\frac{x}{y}=tan\theta$, putting the value we get $(x^2+y^2)^{\frac{3}{2}}=xy$ after this step I am not able to approach further.
 A: No need to eliminate the parameter. Tangents to $(x(t),y(t))$ have slope $\frac{\mathrm dy}{\mathrm dx}$. By the chain rule,
$$\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy}{\mathrm d\theta}\frac{\mathrm d\theta}{\mathrm dx}=\frac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}=\frac{\cos^3\theta-2\sin^2\theta\cos\theta}{2\sin\theta\cos^2\theta-\sin^3\theta}=\frac{1-2\tan^2\theta}{2\tan\theta-\tan^3\theta}$$
The tangent line is parallel to the $x$-axis when the slope is $0$:
$$\frac{1-2\tan^2\theta}{2\tan\theta-\tan^3\theta}=0\implies1-2\tan^2\theta=0\implies\tan\theta=\pm\frac1{\sqrt2}$$
$$\implies\theta=\pm\tan^{-1}\left(\frac1{\sqrt2}\right)+n\pi$$
where $n$ is any integer.
The tangent is parallel to the $y$-axis when the slope approaches $\pm\infty$, or when $\frac{\mathrm dx}{\mathrm dy}=0$:
$$\frac{2\tan\theta-\tan^3\theta}{1-2\tan^2\theta}=0\implies\tan\theta(2-\tan^2\theta)=0\implies\tan\theta=0\text{ or }\tan\theta=\pm\sqrt2$$
$$\implies\theta=2n\pi\text{ or }\theta=\pm\tan^{-1}(\sqrt2)+n\pi$$
The curve has $4$ unique tangents that are either horizontal or vertical when $\theta=\pm\tan^{-1}\left(\frac1{\sqrt2}\right)$ and $\theta=\pm\tan^{-1}(\sqrt2)$, and they have equations $y=\pm\frac2{3\sqrt3}$ and $x=\pm\frac2{3\sqrt3}$. These tangents form the boundary of a square with side lengths $\frac4{3\sqrt3}$, i.e. a square with area $\left(\frac4{3\sqrt3}\right)^2=\color{red}{\frac{16}{27}}$.
