Find the locus of the vertex of a right angle whose two

Find the locus of the vertex of a right angle whose two sides are tangent to the astroid $x=a\cos^3t,\ y=a\sin^3t$.

Step 1: Tangent at parameter $t$

With $\dot x=-3a\cos^2t\sin t,\ \dot y=3a\sin^2t\cos t$, the slope is $y_x'=-\tan t$; the tangent, after multiplying by $\cos t$, is

\[ x\sin t + y\cos t = a\sin t\cos t. \tag{1} \]

Step 2: Perpendicular tangents are a quarter-turn apart

Two tangents at $t_1,t_2$ are perpendicular exactly when $\tan t_1\tan t_2=-1$, i.e.\ $\cos(t_1-t_2)=0$, i.e.\ $t_2=t_1+\pi/2$.

Step 3: Two tangency conditions from $P=(h,k)$

Writing $t$ for $t_1$, and using $\sin(t+\pi/2)=\cos t,\ \cos(t+\pi/2)=-\sin t$ in (1): \[h\sin t+k\cos t = a\sin t\cos t, \qquad h\cos t-k\sin t = -a\sin t\cos t. \tag{2}\]

Step 4: Square and add

\[ h^2+k^2 = 2a^2\sin^2t\cos^2t = \frac{a^2}2\sin^22t. \tag{3} \]

Step 5: Find $\sin t\cos t$ from the sum of the two conditions

Adding the two equations in (2): $(h-k)\sin t+(h+k)\cos t=0$, giving $\tan t=-\dfrac{h+k}{h-k}$, and hence (using $(h-k)^2+(h+k)^2=2(h^2+k^2)$)

\[ \sin t\cos t = \frac{-(h^2-k^2)}{2(h^2+k^2)} \;\Longrightarrow\; \sin^2t\cos^2t = \frac{(h^2-k^2)^2}{4(h^2+k^2)^2}. \tag{4} \]

Step 6: Eliminate the parameter

Substituting (4) into (3) and clearing denominators: \[2(h^2+k^2)^3 = a^2(h^2-k^2)^2. \tag{5}\]

Step 7: Verification

\[a=1.7,\ t=0.3,1.0,1.9,-0.6\text{: intersection }M\text{ of tangents at }t,t+\pi/2\text{ satisfies }2(h^2+k^2)^3-a^2(h^2-k^2)^2\approx0\text{ in all four cases}\ \checkmark\]