I have two parametric equations that I suspect to be equivalent. I know I need to find a bijection map between the two to find whether they are, but I'm not sure how to go about doing so.
The two curves in question are: $$ (t^2,t) $$ and: $$ (2t-\frac{1}{2}(-1\pm\sqrt{8t+1}),2t-\frac{1}{4}(-1\pm\sqrt{8t+1})^2) $$
When plotted, they seem to be identical. Can you give me any pointers as to how I could go about showing their equivalence (or lack of it)?
(As in my commented hint...)
Comparing $P(t)=(t^2, t)$ with
$$Q(t)=(2t-\frac{1}{2}(-1\pm\sqrt{8t+1}),2t-\frac{1}{4}(-1\pm\sqrt{8t+1})^2)$$
means that we want to see whether
$$2t-\frac{1}{2}(-1\pm\sqrt{8t+1})=(2t-\frac{1}{4}(-1\pm\sqrt{8t+1})^2)^2$$
$$=(2t-\frac 14(1\pm 2\sqrt{8t+1}+8t+1))^2$$
$$=4t^2-2t-8t^2\pm (-2)t\sqrt{8t+1}+\frac 14\pm\frac 12\sqrt{8t+1}+4t^2\pm 2t\sqrt{8t+1}+\frac 14(8t+1)+2t$$
$$=2t-\frac 12(-1\pm\sqrt{8t+1})$$
which demonstrates the required equivalence.
However:
(Assuming $t\in\Bbb R$) Note that the domain of $P(t)$ is $(-\infty,+\infty)$, while the domain of $Q(t)$ is $[-\frac 18,+\infty)$, so that the two formulations are only equivalent in terms of how the $x$ and $y$ coordinates relate, not in terms of the domain and range of each parametric form.