If the angle made by the tangent at the point (x_0, y_0) on the curve x=12(t+ t t), y=12(1+ t)^2, 0 t (π)/(2) , with the positive x -axis is (π)/(3) , then y_0 is equal to

Correct answer: C. 27 Given equation of curve x=12(t+ t t), y=12(1+ t)^2 Differentiate w.r.t ' t ', aligned & (d x)/(d t)=12(1+ ^2 t- ^2 t) & (d x)/(d t)=12(1+ 2 t) and (d y)/(d t) & =24(1+ t) t & (d y)/(d x)=(2(1+ t) × t)/(1+ 2 t) aligned

BITSAT 2024Paper 2

Exam: BITSAT
Year: 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Application of Derivatives

BITSAT 2024 · Paper 2 · Mathematics · Application of Derivatives

If the angle made by the tangent at the point (x_0, y_0) on the...

If the angle made by the tangent at the point

\left(x_0, y_0\right)

on the curve

x=12(t+\sin t \cos t), y=12(1+\sin t)^2, 0 \lt t \lt \frac{\pi}{2}

, with the positive x -axis is

\frac{\pi}{3}

, then

y_0

is equal to

Options

A $6(3+2 \sqrt{2})$
B $3(7+4 \sqrt{3})$
C $27$
D $48$

Correct answer

C. $27$

Step-by-step solution

Given equation of curve

x=12(t+\sin t \cos t), y=12(1+\sin t)^2

Differentiate w.r.t '

t

\begin{aligned} & \frac{d x}{d t}=12\left(1+\cos ^2 t-\sin ^2 t\right) \\ & \frac{d x}{d t}=12(1+\cos 2 t) \text{\text{ and } } \frac{d y}{d t} \\ & =24(1+\sin t) \cos t \\ & \frac{d y}{d x}=\frac{2(1+\sin t) \times \cos t}{1+\cos 2 t} \end{aligned}

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