IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ramkumar_november

here is a simple solution :::::::

$p^2=a^2cos^2\theta+b^2sin^2\theta$

$2p^2=a^2(1+cos2\theta)+b^2(1-cos2\theta)$

$2p^2-(a^2+b^2)=(a^2-b^2)cos2\theta$ ...........(1)

differentiating with respect to theta we get

$2p\frac{dp}{d\theta}=(b^2-a^2)sin2\theta$ ........(2)

(1)²+ (2)² .......

$4p^4+(a^2+b^2)^2-4p^2(a^2+b^2)+4p^2\bigg(\frac{dp}{d\theta}\bigg)^2=(a^2-b^2)^2$

on simplification we get ......

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

differentiating with respect to theta.......

$2p\frac{dp}{d\theta}-2\frac{a^2b^2}{p^3}.\frac{dp}{d\theta}+2\bigg(\frac{dp}{d\theta}\bigg).\frac{d^2p}{d\theta ^2}=0$

divide throughout by $2\frac{dp}{d\theta}$ we get

$p-\frac{a^2b^2}{p^3}+\frac{d^2p}{d\theta ^2} = 0$

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

so 1) is the right option.......