Question

The equation of the tangent to the curve $x=2{\cos }^3\left(\theta \right)$ and $y=3{\sin }^3\left(\theta \right)$ at the point, $\theta =\frac{\mathrm{\pi }}{4}$ 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: Solve for the point of contact of tangent with the curve

Given parametric form of the equation is $x=2{\cos }^3\left(\theta \right)$, $y=3{\sin }^3\left(\theta \right)$
The tangent touches the curve at the point where $\theta =\frac{\mathrm{\pi }}{4}$

Therefore, the coordinates of the point of contact is,
$\begin{array}{cccc}& x_{}=2{\cos }^3\left(\frac{\mathrm{\pi }}{4}\right)& & y=3{\sin }^3\left(\frac{\mathrm{\pi }}{4}\right)\\ \Rightarrow & x_{}=2\times {\left(\frac{1}{\sqrt{2}}\right)}^3& \Rightarrow & y=3\times {\left(\frac{1}{\sqrt{2}}\right)}^3\\ \Rightarrow & x_{}=\frac{1}{\sqrt{2}}& \Rightarrow & y=\frac{3}{2\sqrt{2}}\end{array}$ $\left[\because \sin \left(\frac{\pi }{4}\right)=\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\right]$

Thus, the tangent passes through the point and contacts the curve at $\left(\frac{1}{\sqrt{2}},\frac{3}{2\sqrt{2}}\right)$.

Step 2: Solve for the slope of the tangent

The slope of the tangent to a curve is obtained by taking the derivative of the equation of the curve.
Thus, the slope of the tangent to the curve is,

$\begin{array}{c}\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}\\ =\frac{{\displaystyle \frac{d}{d\theta }}\left(3{\sin }^3\left(\theta \right)\right)}{{\displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta }}\left(2{\cos }^3\left(\theta \right)\right)}\\ =\frac{3\left[\left(3{\sin }^2\left(\theta \right)\right)·{\displaystyle \frac{d}{d\theta }}\left(\sin \left(\theta \right)\right)\right]}{2\left[\left(3{\cos }^2\left(\theta \right)\right)·{\displaystyle \frac{d}{d\theta }}\left(\cos \left(\theta \right)\right)\right]}\left[\because \frac{d}{\mathrm{dx}}\left[f\left(g\left(x\right)\right)\right]=f\text{'}\left(g\left(x\right)\right)·g\text{'}\left(x\right)\right]\\ =\frac{9{\sin }^2\left(\theta \right)\cos \left(\theta \right)}{6{\cos }^2\left(\theta \right)\left(-\sin \left(\theta \right)\right)}\left[\because \frac{d}{\mathrm{d\theta }}\left(\sin \left(\theta \right)\right)=\cos \left(\theta \right),\frac{d}{\mathrm{d\theta }}\left(\cos \left(\theta \right)\right)=-\sin \left(\theta \right)\right]\\ \therefore \frac{dy}{dx}=-\frac{3}{2}\tan \left(\theta \right)\mathbf{}\left[\because \tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\right]\end{array}$

The slope at $\theta =\frac{\mathrm{\pi }}{4}$ is,
$\begin{array}{c}{\left(\frac{dy}{dx}\right)}_{\theta =\frac{\mathrm{\pi }}{4}}=-\frac{3}{2}\tan \left(\frac{\mathrm{\pi }}{4}\right)\\ \therefore {\left(\frac{dy}{dx}\right)}_{\theta =\frac{\mathrm{\pi }}{4}}=-\frac{3}{2}\end{array}$

Step 3: Solve for the equation of the tangent

The equation of the tangent can be found by the point-slope form since we know the slope of the tangent and a point it passes through.
The point-slope form of tangent is,
$y-y_1=\left(\frac{dy}{dx}\right)\left(x-x_1\right)$
where $\left(x_1,y_1\right)$ is the coordinates of the point the tangent passes through i.e. $\left(\frac{1}{\sqrt{2}},\frac{3}{2\sqrt{2}}\right)$

Thus the equation of the tangent is,
$\begin{array}{cc}& y-\frac{3}{2\sqrt{2}}=\left(-\frac{3}{2}\right)\left(x-\frac{1}{\sqrt{2}}\right)\\ \Rightarrow & \frac{2\sqrt{2}y-3}{2\sqrt{2}}=-\frac{3x}{2}+\frac{3}{2\sqrt{2}}\\ \Rightarrow & \frac{2\sqrt{2}y-3}{\cancel{2\sqrt{2}}}=\frac{3-3\sqrt{2}x}{\cancel{2\sqrt{2}}}\\ \Rightarrow & 3\sqrt{2}x+2\sqrt{2}y=6\\ \Rightarrow & 3x+2y=3\sqrt{2}\end{array}$

Hence, option(C) is correct.