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If the angle made by the tangent at the point $\mathrm{\left ( x_{0},y_{0} \right )}$ on the curve $\mathrm{x=12\left ( t+\sin t\cos t \right )}$ , $\mathrm{y=12\left ( 1+\sin t \right )^{2},0< t< \frac{\pi}{2}}$ , with the positive x-axis is $\mathrm{\frac{\pi}{3}}$ , then $\mathrm{y_{0}}$ is equal to :
Option: 1
$6\left ( 3+2\sqrt{2} \right )$

Option: 2
$3\left ( 7+4\sqrt{3} \right )$

Option: 3
$27$

Option: 4
$48$

Answers (1)

best_answer

Given,

Curve $\mathrm{x=12\left ( t+\sin t\cos t \right )}$

$\mathrm{y=12\left ( 1+\sin t\right )^{2}}$

Now $\mathrm{\frac{dy}{dx}=\frac{dy/dt}{dx/dt}}$

$\mathrm{ \tan (\pi / 3)=\frac{\frac{d}{d t}\left(12(1+\sin t)^{2}\right)}{\frac{d}{d t}(12(t+\sin t \cos t))} \\ }\\$

$\mathrm{\sqrt{3}=\frac{12 \times 2(1+\sin t) \cos t}{12(1+\cos 2 t)}} \\$

$\mathrm{\sqrt{3}=\frac{2(1+\sin t) \cos t}{2 \cos ^{2} t}}$

$\mathrm{\sqrt{3}=\frac{1+\sin t}{\cos t}=\frac{(\cos t / 2+\sin t / 2)^{2}}{\left(\cos ^{2} t/2-\sin ^{2}t / 2\right)}}\\$

$\mathrm{\sqrt{3}=\tan (t/ 2+\pi / 4)} \\$

$\mathrm{\pi / 3=\frac{t}{2}+\frac{\pi}{4} \Rightarrow \frac{t}{2}=\frac{\pi}{12} \Rightarrow t=\pi / 6}$

Now putting $\mathrm{t=\frac{\pi}{6}}$ in $\mathrm{y=12\left ( 1+\sin t \right )^{2}}$ as $\mathrm{\left ( x_{0},y_{0} \right )}$ lie on curve

So $\mathrm{y_{0}=12\left ( 1+\sin \frac{\pi}{6} \right )^{2}}$

$\mathrm{=12\left ( \frac{3}{2} \right )^{2}=27}$

Hence the correct answer is option 3

Posted by

Gaurav

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