JEE Main 2022 · 25 Jun · Shift 2 · Mathematics Q14

← JEE Main 2022 · 25 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 14 of 90

If the angle made by the tangent at the point $(x_0,y_0)$ on the curve $x=12(t+\sin t\cos t)$ , $y=12(1+\sin t)^2$ , $0<t<\dfrac\pi2$ , with the positive $x$ -axis is $\dfrac\pi3$ , then $y_0$ is equal to:

6\left(3+2\sqrt2\right)

3\left(7+4\sqrt3\right)

Answer and worked solution

Official answer: C

Derivatives. $\dfrac{dx}{dt}=12(1+\cos2t)=24\cos^2t$ , $\dfrac{dy}{dt}=24(1+\sin t)\cos t$ .

Slope. $\dfrac{1+\sin t}{\cos t}=\tan\left(\dfrac\pi4+\dfrac t2\right)=\tan\dfrac\pi3$ , so $t=\dfrac\pi6$ .

Point. $y_0=12\left(1+\frac12\right)^2=27$ .

(C)

Topics:Parametric differentiation Tangents and normals

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JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 14