Find a Parametrization for the following curve? Help?

Question

Find the parametrization for:

The lower half of the parabola $y^2=x-1$

So here's what I did:
I put in $$x=t$$ Then I solved for $y$ $$y^2=t-1$$ $$y=\sqrt {t-1}$$ So the parametrization would be $$x=t$$ $$y=\sqrt {t-1}$$

But for some reason, that doesn't work because if you give $t$ a positive value (for example, $5$), then it would be... $$x=5$$ $$y=2$$ which is wrong because the question asks for the lower half of the parabola $y^2=x-1$, so the $y$ values should be negative.

If you give $t$ a negative value (for example, $5$), then it would be wrong as well... $$x=-5$$ $$y=\text{undefined}$$

Please help? And explain your answer step by step.

Answer 1

Let $y=t$ so $x = t^2 + 1$. If you allow $t$ to take any real value, then you get the entire parabola. To restrict to the lower half, only allow $t \leq 0$.

By the way, if you'd prefer to have a positive parameter—I know I would—then just use $y = -t$ (and $x = (-t)^2 + 1 = t^2 + 1$ doesn't change). This will yield the lower half of the parabola for $t \geq 0$.

Answer 2

I think $x=t$ and $y=-\sqrt{t-1}$ is the parametrizaion you want, because $y=-\sqrt{t-1}$ is a solution of $y^2=t-1$.

Answer 3

The parabola $x=1+y^2$ is symmetric with respect to the $x$ axis and has values $x>1 \quad \forall y \in \mathbb{R}$. So obviously there is no points with $x=-5$.

You parametrization represent the upper half of the parabola, for the lower half you can use the symmetric: $$ x=t \qquad y=-\sqrt{t-1} $$

and note that this is defined (in the sense that $y$ has a real value) for $t\ge 1$.