Question

If $p^2=a^{2}co{s}^2\theta +b^{2}si{n}^2\theta$ then Prove That:

$P+\frac{d^{2}p}{d{\theta }^2}=\frac{a^2{b}^2}{p^3}$

Answer 1

We have,

$p^2=a^2{\cos }^2\theta +b^2{\sin }^2\theta ........\left(1\right)$

On differentiating and we get,

$2p\frac{dp}{d\theta }=a^2(-2\cos \theta \sin \theta )+b^2\left(2\sin \theta \cos \theta \right)$

$2p\frac{dp}{d\theta }=-a^2\sin 2\theta +b^2\sin 2\theta$

$2p\frac{dp}{d\theta }=(-a^2+b^2)\sin 2\theta$

$p\frac{dp}{d\theta }=\frac{(-a^2+b^2)}{2}\sin 2\theta$

$p\frac{dp}{d\theta }=\frac{(-a^2+b^2)}{2}\sin 2\theta .......\left(2\right)$

Again differentiating and we get,

$p\frac{d^{2}p}{d{\theta }^2}+\frac{dp}{d\theta }\frac{dp}{d\theta }=\frac{(-a^2+b^2)}{2}2\cos 2\theta$

$p\frac{d^{2}p}{d{\theta }^2}+{\left(\frac{dp}{d\theta }\right)}^2=\frac{(-a^2+b^2)}{2}2\cos 2\theta$

$p\frac{d^{2}p}{d{\theta }^2}=\frac{(-a^2+b^2)}{2}2\cos 2\theta -{\left(\frac{dp}{d\theta }\right)}^2$

Put the value of $\frac{dp}{d\theta }$ by equation (1)

$p\frac{d^{2}p}{d{\theta }^2}=\frac{(-a^2+b^2)}{2}2\cos 2\theta -{\left(\frac{dp}{d\theta }\right)}^2$

$p\frac{d^{2}p}{d{\theta }^2}=\frac{(-a^2+b^2)}{2}2\cos 2\theta -{\left[\frac{(-a^2+b^2)}{2}\frac{\sin 2\theta }{p}\right]}^2$

$p\frac{d^{2}p}{d{\theta }^2}=\frac{(-a^2+b^2)}{2}2\cos 2\theta -\left[\frac{{(-a^2+b^2)}^2}{4}\frac{{\sin }^{2}2\theta }{p^2}\right]$

$\frac{d^{2}p}{d{\theta }^2}=\frac{\frac{(-a^2+b^2)}{2}2\cos 2\theta -\left[\frac{{(-a^2+b^2)}^2}{4}\frac{{\sin }^{2}2\theta }{p^2}\right]}{p}$

$\frac{d^{2}p}{d{\theta }^2}=\frac{4p^2(-a^2+b^2)\cos 2\theta -{(-a^2+b^2)}^2{\sin }^{2}2\theta }{4p^3}$

L.H.S.

$p+\frac{d^{2}p}{d{\theta }^2}=\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }+\frac{4p^2(-a^2+b^2)\cos 2\theta -{(-a^2+b^2)}^2{\sin }^{2}2\theta }{4p^3}$

On solving that,

$p+\frac{d^{2}p}{d{\theta }^2}=\frac{a^2{b}^2}{p^3}$

Hence, this is the answer..