Finding point on parametric curve such that the tangent is parallel to line

Question

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

I have tried to show it by using mean value theorem. This function is continuous and differentiable on $0\leq \theta \leq\pi/2$. So by Lagrange mean value theorem, there exists $\theta_1; 0\leq \theta_1 \leq\pi/2 $ such that $\left(\frac{dy}{dx}\right)_{ \theta_1}=\frac{y(\pi/2)-y(0)}{a-0}\implies -\tan\theta_1=\frac{a-0}{a-0}=1$. I don't know whether my approach is true or not.

Answer 1

As karakfa notes, you could simply see this by symmetry, but you seem to want a more formal approach.

The parametric derivative is in this case given as $$\frac{3a\cos \theta \sin^2 \theta}{-3a\cos^2 \theta \sin \theta}=-\tan\theta.$$

Since the line between your two points has slope $-1$, we find $\theta$ s.t. $$-\tan\theta=-1 \implies\theta= \frac{\pi}{4}+\pi n, \quad n\in\mathbb{Z},$$

so the solution you are looking for is $$\theta=\frac{\pi}{4}.$$

Answer 2

$$\left(\dfrac{dy/d \theta}{dx/d \theta}\right)_{\theta= \theta_1}=\dfrac{a-0}{0-a}=-1$$