If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$ if $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$

My try :
$2pp' = (b^2-a^2)\sin 2\theta$
$p'^2 + pp'' =  (b^2-a^2)\cos 2\theta$
Thats it ! it doesn't simplify no matter what I try. Any help ?

Since
$$p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta$$
we have
$$p=\mu (a^2 \cos^2 \theta+b^2 \sin^2 \theta)^{1/2}, \mu=\pm 1$$
$$p + p'' =\frac{\mu a^2b^2(\cos^4\theta+\sin^4 \theta+2\cos^2\theta \sin^2\theta)}{(a^2 \cos^2 \theta+b^2 \sin^2 \theta)^{3/2}}=\frac{a^2b^2}{p^3}$$
