Thanks Chris,
What graphing program did you use Chris?
I played with this question too and I found it quite interesting. My results were the same as yours ofcourse. :)
Here are the Desmos graphs,
https://www.desmos.com/calculator/zcpydgu9sb
\(If \quad x=t^2\;\quad y=t+1\qquad \mbox{The restriction is } x\ge0\\ t=\pm\sqrt{x}\qquad t=y-1\\ y-1=\pm \sqrt{x}\\ (y-1)^2=x\\ x=(y-1)^2\\ \)
As Chris said this is a sidways parabola opening in the positive direction with a vertex at (0,1)
Lets look at the second one
a) x=sin^2(t), y=sin(t)+1
The first thing I notice is that t is replaced with sin(t)
so we have
\(x=sin^2t \qquad 0\le x\le 1\\ y=sin(t)+1 \qquad 0\le y \le 2\\ \mbox{So the domain and range are different.}\\ But\;\\ sin^{-1}(\pm\sqrt{x})=t \qquad and \qquad sin^{-1}(y-1)=t\\ \therefore\\ sin^{-1}(\pm\sqrt{x})=sin^{-1}(y-1)\\ \pm\sqrt{x}=(y-1)\\ x=(y-1)^2\\ x=(y-1)^2\\ \mbox{So this is exactly the same as the original}\\ \mbox{ relation except it has different a domain and range} \)
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