Question

If the angle made by the tangent at the point \((x_0, y_0)\) on the curve \(x = 12(t + \sin t \cos t)\), \(y = 12(1 + \sin t)^2\), with \(0<t<\frac{\pi}{2}\), with the positive x-axis is \(\frac{\pi}{3}\), then \(y_0\) is equal to:

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

Hint:

To solve problems involving parametric curves and slopes, always recall the derivative formulas and use the trigonometric identities to simplify expressions.

Answer 1

The Correct Option is C

Step 1: Differentiate the given parametric equations with respect to \(t\) to find \(\frac{dy}{dx}\).
We are given: \[ x = 12(t + \sin t \cos t), \quad y = 12(1 + \sin t)^2 \] \[ \frac{dx}{dt} = 12(1 + \cos^2 t - \sin^2 t), \quad \frac{dy}{dt} = 24(1 + \sin t) \cos t \] \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{24(1 + \sin t) \cos t}{12(1 + \cos^2 t - \sin^2 t)} \] Step 2: Use the condition that the angle between the tangent and the positive x-axis is \(\frac{\pi}{3}\). The slope of the tangent is \( \tan \frac{\pi}{3} = \sqrt{3} \). Thus, equate the slope \(\frac{dy}{dx}\) to \(\sqrt{3}\) and solve for \(t\). \[ \frac{24(1 + \sin t) \cos t}{12(1 + \cos^2 t - \sin^2 t)} = \sqrt{3} \] After solving, we find that \( t = \frac{\pi}{6} \). Substituting this value of \( t \) in \( y = 12(1 + \sin t)^2 \), we get: \[ y_0 = 12(1 + \sin \frac{\pi}{6})^2 = 12 \left(1 + \frac{1}{2}\right)^2 = 12 \times \left(\frac{3}{2}\right)^2 = 27 \] Therefore, \( y_0 = 27 \).