If $(a\cos^3 m, a\sin^3 m)$ is a point on the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$, then find the locus of the intersection point of two tangents that are perpendicular to each other.
My try: I have tried it by taking by two different points. But it got too tedious to proceed. Can anyone please help me giving some easier solution.
On
I'm expanding on the hint provided by @user10354138 :
Let $a=1$. Then $$t\mapsto z(t):=\bigl(\cos^3 t,\sin^3 t\bigr)\qquad(t \ {\rm mod}\ 2\pi)$$ is a parametric representation of the full astroid $\alpha$. Here the variable $t$ is just "time"; it has no geometric significance in the following. Note that $t\mapsto t+{\pi\over2}$ rotates $\alpha$ by ${\pi\over2}$ counterclockwise. It follows that the tangents at $z(t)$ and $z\bigl(t+{\pi\over2}\bigr)$ are orthogonal.
We now fix a $t$. Then the two mentioned tangents $g_i$ have parametric representations $$\eqalign{g_1:\quad u\mapsto w_1(u)&:=z(t)+ u z'(t)\qquad\qquad\qquad(-\infty<u<\infty),\cr g_2:\quad v\mapsto w_2(v)&:=z\bigl(t+{\pi\over2}\bigr)+ v z'\bigl(t+{\pi\over2}\bigr)\qquad(-\infty<v<\infty)\ .\cr}$$ Compute the point of intersection $p:=g_1\wedge g_2$ of these two lines. This point will then be a function of the chosen $t$. In fact $t\mapsto p(t)$ is a parametric representation of the locus we are looking for.
Here is a figure:
I'll give you a hint: use the rotation symmetry of the astroid together with the fact that the slope is monotone on each quadrant to force $m\pm\frac{\pi}{2}$ are the only points whose tangent is perpendicular to that of point cooresponding to parameter $m$.