Given the parametric equations $$\begin{cases}x=at^3-bt\\y=at^3+bt^2\end{cases}$$ If $a$ and $b$ are integers in the interval $[0,100]$, compute the number of ordered pairs $(a,b)$ such that the curve has exactly two distinct $y$-intercepts.
Setting $x=0$ to find all possible values of $t$ that would yield a $y$-intercept, I get $t=0,\pm\sqrt{\frac{b}{a}}$.
$t=0$ would not work because $y$ would then be $0$, which is only one intercept.
Plugging in the other value of $t$, I end up with $y=b\left(\pm\sqrt{\frac{b}{a}}+\frac{b}{a}\right)$. I probably want to figure out which values of $a$ and $b$ generate two different $y$'s, but I'm not sure how to do that.
How could I proceed from here or is there a different approach? Thanks!
If $a=0$, then the graph has only one $y$-intercept ($(0,0)$).
So, $a>0$.
$x=0$ when $t=0$ or $\pm\sqrt{\frac{b}{a}}$.
When $t=0$, $y=0$.
When $t=\sqrt{\frac{b}{a}}$, $y=\frac{b}{a}\left(a\sqrt{\frac{b}{a}}+b\right)=\frac{b\sqrt{b}}{a}\left(\sqrt{a}+\sqrt{b}\right)>0$
When $t=-\sqrt{\frac{b}{a}}$, $y=\frac{b}{a}\left(-a\sqrt{\frac{b}{a}}+b\right)=\frac{b\sqrt{b}}{a}\left(-\sqrt{a}+\sqrt{b}\right)$
The graph has two $y$-intercepts when $a>0$ and $\frac{b\sqrt{b}}{a}\left(-\sqrt{a}+\sqrt{b}\right)=0$, i.e. $a=b$.
So, $(a,b)=(1,1),(2,2),\dots,(100,100)$.