<s>[INST] 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 ?
 [/INST] We need to establish  $$p+p''=\frac{a^2b^2}{p^3}$$
Multiplying by $2p'(\ne0),$ $$(p+p'')2p'=a^2b^2\cdot\frac{2p'}{p^3}$$
Integrating we get,
$$p^2+p'^2=-\frac{a^2b^2}{p^2}+K\text{, where }K\text{ is an arbitrary constant} $$
Multiplying by $p^2(\ne0),$
$$p^4+p'^2p^2=-a^2b^2 +p^2\cdot K\  \  \  \ (1)$$
Now, $$p^4+p'^2p^2=(p^2)^2+(pp')^2=(a^2\cos^2\theta+b^2\sin^2\theta)^2+[(b^2-a^2)\sin\theta\cos\theta]^2$$
$$=a^4\cos^2\theta(\cos^2\theta+\sin^2\theta)+b^4\sin^2\theta(\cos^2\theta+\sin^2\theta)$$
$$=a^2(p^2-b^2\sin^2\theta)+b^2(p^2-a^2\cos^2\theta)$$
$$\implies p^4+p'^2p^2=p^2(a^2+b^2)-a^2b^2$$
Comparing with $(1),K=a^2+b^2$ which definitely a constant independent of $\theta$
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