# Let C be a curve defined parametrically as

Question:

Let $C$ be a curve defined parametrically as $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta, 0 \leq \theta \leq \frac{\pi}{2}$. Determine a point $\mathrm{P}$ on $C$, where the tangent to $C$ is parallel to the chord joining the points $(a, 0)$ and $(0, a)$

Solution:

As, $x=a \cos ^{3} \theta$

$\Rightarrow \frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta$

And, $y=a \sin ^{3} \theta$

$\Rightarrow \frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta$

So, $\frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}=-\tan \theta$

For the tangent to be parallel to the chord joining the points $(a, 0)$ and $(0, a)$,

$\frac{d y}{d x}=\frac{0-a}{a-0}$

$\Rightarrow-\tan \theta=-1$

$\Rightarrow \tan \theta=1$

$\Rightarrow \theta=\frac{\pi}{4}$

Now,

$x=a \cos ^{3} \frac{\pi}{4}=a\left(\frac{1}{\sqrt{2}}\right)^{3}=\frac{a}{2 \sqrt{2}}$ and

$y=a \sin ^{3} \frac{\pi}{4}=a\left(\frac{1}{\sqrt{2}}\right)^{3}=\frac{a}{2 \sqrt{2}}$

So, the point $\mathrm{P}$ on the curve $C$ is $\left(\frac{\mathrm{a}}{2 \sqrt{2}}, \frac{\mathrm{a}}{2 \sqrt{2}}\right)$.