If p1  and p2  are lengths of the perpendicular from origin on the tangent and normal to the curve x2/3+y2/3=62/3  then the value of 4p12+p22  is equal to

Any point on the curve x2/3+y2/3=62/3  is of the form (6cos3θ,6sin3θ). Now dy/dx=−y1/3/x1/3,  so dydx|(6cos3θ,6sin3θ)=−sinθcosθ=−tanθ. Thus the equation of tangent at (6cos3θ,6sin3θ). Y−6sin3θ=−tanθ(X−6cos3θ) ⇒    Xsinθ+Ycosθ=6sin3θcosθ+6sinθcos3θ=6sinθcosθ. The equation of normal at (6cos3θ,6sin3θ)  isY−6sin3θ=cotθ(X−6cos3θ), i.e.  cosθX−sinθY=6(sin2θ−cos2θ)                                 =−6cos2θ Therefore , p1=6sinθcosθ  and  p2=6cos2θ                    4p12+p22=36sin22θ+36cos22θ=36.