Find the locus of the intersection point of two perpendicular tangents to the curve $xy^2=1$.
I find out that the tangents to this curve with slope $m$ has this general form:
$y = mx+(-2m)^{\frac{1}{3}}-m \left( -\frac{1}{2m} \right)^{\frac{2}{3}}$
The perpendicular tangent would have this form then:
$y = -\frac{1}{m}x+ \left(\frac{2}{m} \right)^{\frac{1}{3}}+\frac{1}{m} \left( \frac{m}{2} \right)^{\frac{2}{3}}$
I solved geometrically that the locus is the circle: $2x^2+2y^2-3\sqrt[3]{2}x=0$ (you can check it here: desmos.com/calculator/xt2xquh72l), but I couldn't solve it algebraically. I wonder if you could help me in eliminating $m$ in this system:
$\begin{cases} y &= mx+(-2m)^{\frac{1}{3}}-m \left( -\frac{1}{2m} \right)^{\frac{2}{3}} \\ y &= -\frac{1}{m}x+ \left(\frac{2}{m} \right)^{\frac{1}{3}}+\frac{1}{m} \left( \frac{m}{2} \right)^{\frac{2}{3}} \\ \end{cases} $