JEE Main 2022 (Online) 25th June Evening ShiftQuestion 1
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:
Differentiation and its Application
Solution
Concept. The slope is $\frac{dy/dt}{dx/dt}$; set it equal to $\sqrt3$ and solve for $\sin t$. Formulas used. (1) $1+\cos2t=2\cos^2t$Step 1 - Derivatives. By (1), $$\frac{dx}{dt}=12(1+\cos2t)=24\cos^2t,\qquad\frac{dy}{dt}=24(1+\sin t)\cos t.$$ Step 2 - Slope. $\frac{dy}{dx}=\frac{1+\sin t}{\cos t}=\sqrt3$. Squaring: $(1+s)^2=3(1-s^2)$, and dividing by $1+s$: $1+s=3(1-s)$, so $s=\sin t=\frac12$. Step 3 - $y_0$. $y_0=12\left(\frac32\right)^2=27$. Answer: (C)