Curvature and conjugate diameter

Question

If P and P' be thr radii of curvature at extremities of two conjugate diameters of an ellipse x^2/a^2 + y^2/b^2=1 

 Then prove that  

       (P^2/3  + P'^2/3) ( ab)^2/3 =a^2 + b^2 .

Answer 1

Radius of curvature for ellipse: P=b^2/a, P'=a^2/b. Hence,

 

(P^2/3+P'^2/3)(ab)^2/3=((b^2/a)^2/3+(a^2/b)^2/3)(ab)^2/3=

=(b^(4/3+2/3)+a^(4/3+2/3)=a^2+b^2.

 b²/a