UPSC NDA Math Model Paper 10

Given: The equation of curve is x = sin3t, y = cos2t. Here we have to find the equation of the tangent the curve x = sin3t, y = cos2t at t = π/4 When t = π/4, then x = sin (3π/4) = sin [π - (π/4)] = sin π/4 $= \frac{1}{\sqrt{2}}$ Similarly, when t = π/4, then y = cos [2 ⋅ (π/4)] = cos π/2 = 0 So, the point of contact is $\left(\frac{1}{\sqrt{2}}, 0\right)$ As we know that slope of the tangent at any point say $(x_1, y_1)$ is given by: $m = \left[\frac{dy}{dx}\right]_{(x_1, y_1)}$ Now by differentiating x = sin3t, y = cos2t with respect to t we get, ⇒ dx/dt = 3 cos 3t and dy/dt = - 2 sin 2t As we know that $\Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{g'(t)}{f'(t)}$ and $\frac{dx}{dy} = \frac{\frac{dx}{dt}}{\frac{dy}{dt}} = \frac{f'(t)}{g'(t)}$ $\Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -\frac{2 \sin 2t}{3 \cos 3t}$ $\Rightarrow \left[\frac{dy}{dx}\right]_{t=\frac{\pi}{4}} = -\frac{2 \sin\left(\frac{\pi}{2}\right)}{3 \cos\left(\frac{3\pi}{4}\right)}$ $= \frac{2 \sin\left(\frac{\pi}{2}\right)}{3 \cos\left(\frac{\pi}{4}\right)} = \frac{2\sqrt{2}}{3}$ As we know that equation of tangent at any point say $(x_1, y_1)$ is given by: $y - y_1$ $= \left[\frac{dy}{dx}\right]_{(x_1, y_1)} \cdot (x - x_1)$ Here, $x_1 = \frac{1}{\sqrt{2}}, y_1 = 0$ and $\left[\frac{dy}{dx}\right]_{t=\frac{\pi}{4}} = \frac{2\sqrt{2}}{3}$ $\Rightarrow y - 0 = \frac{2\sqrt{2}}{3} \cdot \left(x - \frac{1}{\sqrt{2}}\right)$ $\Rightarrow 2\sqrt{2} x - 3y - 2 = 0$ Hence, equation of the required tangent is $2\sqrt{2} x - 3y - 2 = 0$