Finding point on parametric curve such that the tangent is parallel to line 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.
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 $$\theta=\frac{\pi}{4}.$$
From the given coordinates, the slope can be calculated to be = -1.

Now as in previous answers I can see that θ=π/4 has been calculated and well shown.

This value can be now be used in x=acos^3θ, where x=a/2.2^(1/2).Similarly, y(whenθ=π/4)= a/2.2^(1/2).

Those will be the answer to the requisite point on C.