Parametrize the "alpha curve" I wanted to check if my solution would be correct for the question:
Q. Parametrize the "alpha curve"
$y^2$=$x^3$+$x^2$
A. y=tx
$(tx)^2$=$x^3$+$x^2$
$t^2$$x^2$=$x^3$+$x^2$
$t^2$$x^2$-$x^2$=$x^3$
$x^2$($t^2$-1)=$x^3$
x=$t^2$-1
Substitute x=$t^2$-1 into $y^2$=$x^3$+$x^2$
$y^2$=$x^3$+$x^2$
= $x^2$(x+1)
=($t^2$-1)$^2$($t^2$-1+1)
y=${\sqrt{(t^2-1)^2(t^2)}}$
=$t(t^2-1)$
So, the parametrization of the alpha curve is
$t\mapsto(t^2-1, t(t^2-1)$
and the double point (0,0) is obtained twice for t=${\sqrt{1}}$

In polar coordinates,
$$r^2\sin^2(\theta)=r^3\cos^3(\theta)+r^2\cos^2(\theta),$$
then
$$r=\frac{\sin^2(\theta)-\cos^2(\theta)}{\cos^3(\theta)}=\frac{1-2\cos^2(\theta)}{\cos^3(\theta)}.$$
Then numerator cancels for the four angles $\theta=\pm\frac\pi4,\pm\frac{3\pi}4$, corresponding to a double point at the origin, the curve being tangent to the main bissectors. The denominator cancels for $\theta=\frac\pi2,\frac{3\pi}2$, giving points at infinity.
