If the angle made by the tangent at the point x0,y0 on the curve x=12(t+sin t cos t),y=12(1+sint)2,0<t<π2, with the positive x-axis is π3 then y0 is equal to

dydx=dydtdxdt=2(1+sint)×cost1+cos2t ⇒2(1+sint)cost2cos2t=3 ⇒t=π6 ⇒y0=27

If the angle made by the tangent at the point x0,y0 on the curve x=12(t+sin t cos t),y=12(1+sint)2,0<t<π2, with the positive x-axis is π3 then y0 is equal to

dydx=dydtdxdt=2(1+sint)×cost1+cos2t ⇒2(1+sint)cost2cos2t=3 ⇒t=π6 ⇒y0=27

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If the angle made by the tangent at the point $(x_{0}, y_{0})$ on the curve $x = 12 (t + \sin t \cos t),$
$y = 12 (1 + \sin t)^{2}, 0 < t < \frac{π}{2}$ , with the positive x-axis is $\frac{π}{3}$ then y₀ is equal to

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$6 (3 + 2 \sqrt{2})$

$3 (7 + 4 \sqrt{3})$

answer is C.

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$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 (1 + \sin t) \times \cos t}{1 + \cos 2 t} \Rightarrow \frac{2 (1 + \sin t) \cos t}{2 \cos^{2} t} = \sqrt{3} \Rightarrow t = \frac{π}{6} \Rightarrow y_{0} = 27$

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If the angle made by the tangent at the point x0,y0 on the curve x=12(t+sin t cos t),y=12(1+sint)2,0<t<π2, with the positive x-axis is π3 then y0 is equal to

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If the angle made by the tangent at the point $(x_{0}, y_{0})$ on the curve $x = 12 (t + \sin t \cos t),$
$y = 12 (1 + \sin t)^{2}, 0 < t < \frac{π}{2}$ , with the positive x-axis is $\frac{π}{3}$ then y₀ is equal to