Finding the $y$-intercept of a parametric equation

Question

Find the $y$-intercept of a given parametric curve.

$$x=t(t^2-3)$$

$$y=3(t^2-3)$$

My attempt: $x'_t=3t^2-3$, and $y'_t=6t$

$y'/x'=6t/3t^2-3=2t/t^2-1=?$

And that's where I'm stuck, I tried the other method of substitution but couldn't find a method to isolate $t$. I derived $x$ and $y$ because I'm trying to find $mx+b$ to find its y-intercept.

Answer 1

The $y$ intercept is when $x=0$, so solve the $x$ equation to get the $t$ that makes that true. You should find three of them. Each creates a $y$ intercept, but two are the same.

Here is the plot from Alpha:

You can see it crosses the $y$ axis once at $-9$ and twice at $0$. These correspond to $t=0$ and $t=\pm \sqrt 3$

Answer 2

The $y$ intercepts of a curve are the points at which the curve intersects the $y$-axis (i.e. the curve may be a line and may be not, so the $y$-intercept is not only the $b$ in the equation $y=mx+b$)

Also when the curve intersects the $y$-axis, it means that the $x=0$ so $t(t^2-3)=0 \implies t=0$ or $t=\pm \sqrt{3}$

Your graph will look like this

Answer 3

We have x=t(t^2-3) and y=3(t^2-3) So x/y = t/3 => t=3x/y So substituting this value of t in any of the parametric expression in t for x or y we get an equation in x and y. Further we can put x=0 to get the y intercept.