I want to parametrize the curve $y^2=x^2+x^3.$ When I was reading the textbook by Theodore Shifrin "DIFFERENTIAL GEOMETRY:A First Course in Curves and Surfaces". I found the parametrisation as $x(t)=t^2-1, y(t)=t(t^2-1). $ Why did the author used the intersection of the line $y = tx$ with the curve to find the parametrisation?
My attempt:-
We have $y^2=x^2+x^3$. Put $y=tx$, we get $(tx)^2 = x^2(x + 1).$
That is $t^2x^2-x^2.x-x^2 = 0\implies x^2(t^2-x-1)=0.$ That is either $x=0$ or $x=t^2-1$. Then author assumed $t\ge 0$, the got $y=t(t^2-1)$.
Can I parametrize in different way? When I tried to parametrize using $x=t$. I am not getting a functional relation for $y$ which covers the curve.
Could you help me?
The author did not come up with this by divine inspiration. This is a classical construction, going back hundreds of years. The idea is this: The curve $y^2=x+x^3$ is a cubic curve, so you expect a line to intersect it in three points. (Draw some pictures and verify this for yourself.) The origin is a double point of the curve (you can see that two "branches" cross there) and so lines passing through the origin will intersect the curve at a unique point other than the origin — the origin uses up two of the three points.