Question

If a curve is given by $ x=a\cos t+\frac{b}{2}\cos 2t $ and $ y=\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{2{{a}^{2}}+{{b}^{2}}}{3ab} $
• B. $ \cos t=\frac{{{a}^{2}}+2{{b}^{2}}}{3ab} $
• C. $ \tan t=a/b $
• D. None of the above

Answer 1

Answer: (B)

We have, $x=a \cos t+\frac{b}{2} \cos 2 t$ and
$y=a \sin t+\frac{b}{2} \sin 2 t$ On-differentiating both
equations w.r.t.t, we get $\frac{d x}{d t}=-a \sin t-b \sin 2 t$ and
$\frac{d y}{d t}=a \cos t+b \cos 2 t \therefore \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a \cos t+b \cos 2 t}{-a \sin t-b \sin 2}$
$\Rightarrow $
$\frac{d y}{d x}=-\frac{(a \cos t+b \cos 2 t)}{(a \sin t+b \sin 2 t)}$ Now, $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d t}\left(\frac{d y}{d x}\right)$
$\cdot \frac{d t}{d x} \Rightarrow $
$\frac{d^{2} y}{d x^{2}}=-\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d t}\left(\frac{d y}{d x}\right) \cdot \frac{d t}{d x}$
$=\frac{\left[\begin{array}{l}(a \sin t+b \sin 2 t)(-a \sin t-2 b \sin 2 t) \\ -(a \cos t+b \cos 2 t)(a \cos t+2 b \cos 2 t)\end{array}\right]}{\left[\begin{array}{l}a^{2} \sin ^{2} t+3 a b \sin t \sin 2 t+2 b^{2} \sin ^{2} 2 t \\ (a \sin t+b \sin 2 t)^{2}\end{array}\right]}$ $\times \frac{1}{-a \sin t-b \sin 2 t}=\frac{\frac{d a b \cos ^{2} t+3 a b \cos t \cos 2 t+2 b^{2} \cos ^{2} 2 t}{(a)}}{a^{2}\left(\sin ^{2} t+\cos ^{2} t\right)+b^{2}\left(\sin ^{2} 2 t+\cos ^{2} 2 t\right)}$ $\quad+3 a b \cos (2 t-t)$ $=\frac{(a \sin t+b \sin 2 t)^{3}}{\Rightarrow a^{2}+2 b^{2}+3 a b \cos t=0 \Rightarrow \cos t}=$ $-\left(\frac{d^{2} y}{d x^{2}}=-\left[\frac{a^{2}+2 b^{2}+3 a b \cos t}{(a \sin t+b \sin 2 t)^{3}}\right]\right.$ Given, $\left.\frac{d^{2} y}{d x^{2}}=0 \Rightarrow a^{2}\right)$