Parametrize the "alpha curve"

Question

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=${\sqrt{1}}$

Answer 1

The fast way:

$$\left(\frac yx\right)^2=x+1.$$

Set $\dfrac yx=t$ so that

$$\begin{cases}x=t^2-1,\\y=tx=t(t^2-1).\end{cases}$$

Answer 2

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.