Question

Pre-Calculus? Parametric Functions?

Answer 1

Notice that if you just plug in
`x=t` into the equation for
`y`, you end up with


`\begin{cases}x=t\\y=3t^2-4\end{cases}\implies y=3x^2-4`

You have to do the same thing with the given choices. For example, if


`\begin{cases}x=2t\\y=12t^2-4\end{cases}`

then we see that
`t=\frac x2`, and plugging this into the second equation gives


`y=12\left(\frac x2\right)^2-4=\frac{12x^2}4-4=3x^2-4`

which matches the original set of parametric equations.

So the general strategy is to eliminate the parameter
`t` by solving for it in each
`x(t)` equation. Then substitute this result for the
`t` in the corresponding
`y(t)` equation, and see if it reduces to the same equation at the top.