Question

lf a curve is given by $x=a\cos t+\frac{b}{2}\cos 2t$ and $y=a\sin t+\frac{b}{2}\sin 2t$, then the points for which$\frac{d^{2}y}{d{x}^2}=0$ are given by

• A. $\sin t=\frac{2a^2+b^2}{3ab}$
• B. $\cos t=-\frac{a^2+2b^2}{3ab}$
• C. $\tan t=\frac{a}{b}$
• D. $\cot t=\frac{a}{b}$

Answer 1

Answer: (B)

$\cos t=-\frac{a^2+2b^2}{3ab}$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{acost+bcos2t}{-asint-bsin2t}=-cott+\frac{b}{sint(a+2bcost)}$

$\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\frac{dy}{dx}}{\frac{dx}{dt}}=\frac{1}{si{n}^{2}t}-\frac{b(acost+2bcos2t)}{(asint+bsin2t{)}^2}$

GIven $\frac{d^{2}y}{d{x}^2}=0$

$\Rightarrow a^{2}si{n}^{2}t+3absi{n}^{2}tcost+2b^{2}si{n}^{2}t=0$

$\Rightarrow cost=-\frac{a^2+2b^2}{3ab}$

So, Option (B)