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

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Differentiation

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

y_{0}

is equal to

Answer: (C)

For parametric curves,

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

. Here

\frac{d x}{d t} = 12 (1 + cos 2 t) = 24 cos^{2} t

and

\frac{d y}{d t} = 24 (1 + sin t) cos t

. So

\frac{d y}{d x} = \frac{24 ( 1 + s i n t ) c o s t}{24 c o s ^{2} t} = \frac{1 + s i n t}{c o s t}

. The tangent makes angle

\frac{π}{3}

, so

\frac{d y}{d x} = tan \frac{π}{3} = 3

. Thus

\frac{1 + s i n t}{c o s t} = 3

, giving

1 + sin t = 3 cos t

. Squaring:

1 + 2 sin t + sin^{2} t = 3 cos^{2} t = 3 (1 - sin^{2} t)

, so

4 sin^{2} t + 2 sin t - 2 = 0

, or

2 sin^{2} t + sin t - 1 = 0

. Solving:

sin t = \frac{- 1 \pm 3}{4}

. Since

0 < t < \frac{π}{2}

sin t = \frac{1}{2}

, so

t = \frac{π}{6}

. Then

y_{0} = 12 (1 + \frac{1}{2})^{2} = 12 \cdot \frac{9}{4} = 27

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option