Prove that a parametric equation's range is subset of a cartesian equation

Question

$r(t)=t(t-2)^3i +t(t-2)^2j$ where $r:\Bbb R \to \Bbb R^2$

$$C = \{(x,y)∈\Bbb R^2 \mid y^4=x^3+2x^2y\}$$

I have difficulty with this question. This is where I am up to.

1. Find $dy/dx$ of the cartesian curve $dy/dx = (3x^2+4xy)/(4y^3-2x^2)$;
2. $(4y^3-2x^2) = 0, y= (2x^{2/3})/2$;
3. Substitute $y$ into cartesian equation and find $x=-27/16, 0$.

I'm not sure where to go from here, can anyone clarify if I am on the right track, thanks!

Answer 1

I think you're making it too complicated.

If $x(t)=t(t-2)^3$ and $y(t)=t(t-2)^2$,

then it's easy to show that $y^4=x^3+2x^2y$:

$t^4(t-2)^8=t^3(t-2)^9+2t^2(t-2)^6t(t-2)^2$

because the right side is $t(t-2)^8[t^2(t-2)+2t^2]=t(t-2)^8[t^3].$