Any point on the curve $x^{2/3}+y^{2/3}=6^{2/3}$  is of the form

$\left(6{\cos }^3\theta ,6{\sin }^3\theta \right).$

Now $dy/dx=-y^{1/3}/x^{1/3},$  so ${\frac{dy}{dx}|}_{\left(6{\cos }^3\theta ,6{\sin }^3\theta \right)}=-\frac{\sin \theta }{\cos \theta }=-\tan \theta .$

Thus the equation of tangent at $\left(6{\cos }^3\theta ,6{\sin }^3\theta \right).$

$Y-6{\sin }^3\theta =-\tan \theta \left(X-6{\cos }^3\theta \right)$

$\Rightarrow X\sin \theta +Y\cos \theta =6{\sin }^3\theta \cos \theta +6\sin \theta {\cos }^3\theta =6\sin \theta \cos \theta .$

The equation of normal at $\left(6{\cos }^3\theta ,6{\sin }^3\theta \right)$  is

$Y-6{\sin }^3\theta =\cot \theta \left(X-6{\cos }^3\theta \right),$

$i.e.\cos \theta X-\sin \theta Y=6\left({\sin }^2\theta -{\cos }^2\theta \right)$

                                $=-6\cos 2\theta$

Therefore , $p_1=6\sin \theta \cos \theta$  and  $p_2=6\cos 2\theta$

              $4p_1{}^2+p_2{}^2=36{\sin }^{2}2\theta +36{\cos }^{2}2\theta =36.$