If p and q are lengths of perpendiculars from the origin on the tangent and the normal to the curve $x^{2/3}+y^{2/3}=a^{2/3},$ then $4p^2+q^2=$

$x=a{\cos }^3\theta ,y=a{\sin }^3\theta$

$\frac{dy}{dx}=\frac{-\sin \theta }{\cos \theta }$

Equation of tangent 

$y-a{\sin }^3\theta =\frac{-\sin \theta }{\cos \theta }\left(x-a{\cos }^3\theta \right)$

$x\sin \theta +y\cos \theta =a\cos \theta \sin \theta$

P=length of ${\perp }^{el}$ from $\left(0,0\right)$ is 

$\frac{\left|c\right|}{\sqrt{a^2+b^2}}=\frac{\left|a\sin \theta \cos \theta \right|}{\sqrt{{\sin }^2\theta +{\cos }^2\theta }}$

$P=\frac{1}{2}a.\sin 2\theta$

Equation of normal 

$y-a{\sin }^3\theta =\frac{\cos \theta }{\sin \theta }\left(x-a{\cos }^3\theta \right)$

$x\cos \theta -y\sin \theta =a\cos 2\theta$

Q=length of ${\perp }^{er}$ from $\left(0,0\right)$ is 

$\frac{\left|-a\cos 2\theta \right|}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}$

$\therefore q=a\cos 2\theta$

$4p^2+q^2=\cancel{4}.\frac{1}{\cancel{4}}.a^2{\sin }^{2}2\theta +a^2{\cos }^{2}2\theta =a^2\left(1\right)=a^2$