Parametric equations, eliminating the parameter $\,x = t^2 + t,\,$ $y= 2t-1$

Question

$$x = t^2 + t\qquad y= 2t-1$$

So I solve $y$ for $t$

$$t = \frac{1}{2}(y+1)$$

Then I am supposed to plug it into the equation of $x$ which is where I lose track of the logic.

$$x = \left( \frac{1}{2}(y+1)\right)^2 + \frac{1}{2}(y+1) = \frac{1}{4}y^2 + y+\frac{3}{4}$$

That is now my answer? I am lost. This is x(y)? How is this valid? I don't understand.

Answer 1

The variable $t$ in the parametized equations is "the same" $t$: Both $x$ and $y$ are defined in terms of the same variable $t$.

So in solving for $t$ in terms of $y$: $t = \frac 12(y + 1)$, we can use this "definition" of $t$ by substituting it into the equation for $x$: $$x = t^2 + t$$ given $t = \frac 12(y+ 1)$ gives us $$x = \frac 14 y^2 + y + \frac 34$$ as you show.

This then gives us a function of $x$ in terms of $y$: indeed, $x(y) = \frac 14 y^2 + y + \frac 34$ is a "horizontal parabola".

Answer 2

As $t$ varies through $\mathbb R$, $y$ varies through $\mathbb R$ and $x$ varies too though not through all of $\mathbb R$. The equation you have obtained expresses $x$ in terms of $y$ thus omitting the use of the parameter $t$.

Answer 3

Let's assume you are walking on an xy-plane. Your x-position (or east-west position) at a certain time t is given by $x = t^2 + t$, and your y-position is $y = 2t - 1$.

If you want to know what the whole path you remained is, without wanting to know when you stepped on where? Eliminate t:

$$x = \frac{1}{4} y^2 + y + \frac{3}{4}$$

If a pair of $(x, y)$ satisfied that, that means some time in the past or in the future, you stepped or would step on that point.