Question

If $p^2=a^2{\cos }^2\theta +b^2{\sin }^2\theta$ then $\frac{d^{2}p}{d{\theta }^2}+p$ is equal to $(a\ne b)$

• A. $\frac{a^2{b}^2}{p^4}$
• B. $\frac{a^2{b}^2}{p^2}$
• C. $\frac{ab}{p}$
• D. $\frac{a^2{b}^2}{p^3}$

Answer 1

Answer: (D)

$\frac{a^2{b}^2}{p^3}$

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