# Thread: Help with Parametric Equation

1. ## Help with Parametric Equation

I'm an Algebra 2 student and we are learning parametric equations. I am having difficulty solving this one, although I'm good on most of the others.

Here it is:

y = 2t
x = t^2 + 2t

The trouble I'm having is coming up with a "y=" equation, because the y term becomes quadratic. What I have gotten down to is: y^2 = 4x - 4y. Where do I go from here? Thanks...

2. An easy way of solving this is solving for t on the x=t^2+2t equation and putting into the other equation y=2t
you have to solve it by completing the square on x=t^2+2t
so after doing that, you get x+1=(t+1)^2
then square root both sides +√(x+1)=t+1 and -√(x+1)=t+1
then -1 to both sides t=-1+√(x+1) and t=-1-√(x+1)
substitute both of those into the equation y=2t
and you get y=2(-1+√(x+1)) and y=2(-1-√(x+1))
simplified, you get y=-2+2√(x+1) and y=-2-2√(x+1)

you get two equations because when you graph the parametric equation, you get a sideways parabola, therefore, it is not a function, and has to be split into two different functions so it is possible to be graphed.