In a recent quiz (answers have since been released, etc), my GSI gave the problem
Write a parameterization of the curve $$y^2 = x^2(x-1)$$ in terms of some parameter $t$. (Hint: consider the line of slope $t$ through the origin. How many times does this line intersect the curve?)
Before we took the quiz, he changed the problem to finding the parameterization of the curve $$y^2=x^2(x+1)$$ to make it easier. Later, in discussion when going over the quiz, he gave the caveat that the problem was too hard, and said that the goal was to take the hint and realize we could use $y=tx$ since that intersected the curve, a cubic, at three points, and then from there plug it back into the formula for $y$ and obtain the full parameterization $x=t^2-1, y=t(t^2-1)$ (I am summarizing his approach slightly as I frankly didn't entirely follow.)
I have a few related questions about this scenario.
What is the intuition behind realizing $y=tx$ works? He said this was basically given to us in the hint as a starting point, but say there was no hint - what is the idea behind this? I think my lack of understanding of this is a big reason why I don't follow the problem. It's also supposed to be a broader technique, since part c) of the same problem asked us to generalize the process we used in the above problem to any curve $y^2 = f(x)$ where $f(x)$ is a cubic polynomial with multiple roots.
Once one realizes $y=tx$ works, he said we could use the fact it was a cubic to know that points of intersection had to be either the origin or $x\neq 0$, allowing us to divide out $x^2$ in the substitution of $tx$ for $y$ and get the parameterization $x = t^2-1$. What exactly does this mean? I understand that a cubic will have three solutions (whether real or imaginary) but I'm not sure how that leads to this line of argument (I think I may be misunderstanding his process/misphrasing his explanation).
How would this be approached if the original problem is kept, and it was $(x-1)$ instead of $(x+1)$? How does the latter make it easier?
Thank you!