Question

The equation of the tangent to the curve $x=2{\cos }^3\theta$ and$y=3{\sin }^3\theta$ at the point$\theta =\frac{\pi }{4}$ is

• A. $2x+3y=3\sqrt{2}$
• B. $2x-3y=3\sqrt{2}$
• C. $3x+2y=3\sqrt{2}$
• D. $3x-2y=3\sqrt{2}$

Answer 1

The correct option is C

$3x+2y=3\sqrt{2}$

Explanation for the correct option:

Step 1: Find the first order derivative of the equation of the given curve

Given, Equation of curve $x=2{\cos }^3\theta$ and $y=3{\sin }^3\theta$

We are given, $x=2{\cos }^3\theta$ and $y=3{\sin }^3\theta$

$\therefore \frac{dx}{d\theta }=-6{\cos }^2\theta \sin \theta$ and $\frac{dy}{d\theta }=9{\sin }^2\theta \cos \theta$

$$\begin{gathered} \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }} \\ =-\frac{3}{2}\tan \theta \left[\because \frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}=\mathrm{tan\theta }\right] \end{gathered}$$

Step 2:Find the slope of the tangent at the required point

At $\theta =\frac{\mathrm{\pi }}{4}$,

$x=2{\cos }^3\frac{\mathrm{\pi }}{4}$ and $y=3{\sin }^3\frac{\mathrm{\pi }}{4}$

$x=\frac{1}{\sqrt{2}}$ and $y=\frac{3}{2\sqrt{2}}$

The first order derivative at a given point gives the slope of the tangent at that point

$\therefore {\left(\frac{dy}{dx}\right)}_{\left(\theta =\frac{\mathrm{\pi }}{4}\right)}=\frac{-3}{2}$

Step 3: Find the equation of the tangent using slope-point form

Therefore, equation of tangent of the curve passing through $\left(\frac{1}{\sqrt{2}},\frac{3}{2\sqrt{2}}\right)$ is

$$\begin{gathered} y-\frac{3}{2\sqrt{2}}=-\frac{3}{2}\left(x-\frac{1}{\sqrt{2}}\right) \\ \Rightarrow y-\frac{3}{2\sqrt{2}}=-\frac{3}{2}x+\frac{3}{2\sqrt{2}} \\ \Rightarrow \frac{3}{2}x+y=\frac{3}{2\sqrt{2}}+\frac{3}{2\sqrt{2}} \\ \Rightarrow \frac{3x+2y}{2}=\frac{6}{2\sqrt{2}} \\ \Rightarrow 3x+2y=\frac{6}{\sqrt{2}} \\ \Rightarrow 3x+2y=3\sqrt{2} \end{gathered}$$

Hence, the correct answer is option (C).