Find the plot of $y=1+\cos t$, $x=\sin^2t$. I'm trying to find the plot for the following :
$$y=1+\cos t, x=\sin^2t$$
I'm trying to get ride off variable $t$.
This is what I done for some reason is incorrect :
$$x=\sin^2t=\frac{1}{2}-\frac{\cos 2t}{2}=\frac{1}{2}-\frac{2\cos^2t-1}{2}=-\cos^2t+1$$
We know that :
$y=1+\cos t \rightarrow y-1=\cos t$
Therefore
$$x=-\cos^2t+1=-(y-1)^2+1 \rightarrow x=-(y-1)^2+1$$
But for some reason I get the wrong plot.
Any ideas?
Thank you!
 A: you have the parametric equation $$y = 1 + \cos t, x = \sin^2t.$$  eliminating $t$ using the fact $\sin^2 t + \cos ^2 t = 1,$ we get  $$(y-1)^2 + x = 1 $$ the graph is a parabola in the first quadrant with axis of symmetry at $y = 1,$ with vertex at $(1,1).$ 
A: As you have already obtained, the third parametric variable $t$ can be easily eliminated by substituting $\cos t=y-1$ in the second relation $x=\sin^2 t$, we get $$x=1-\cos^2 t=1-(y-1)^2 \implies (y-1)^2=-(x-1)$$ Now, compare the above equation with the standard equation of the parabola $Y^2=-4AX$ (having X-axis as its axis & sides opening in negative X-direction), we have $Y=y-1$, $x=x-1$ & $A=\frac{1}{4}$. Thus, the parabola has its vertex is at $V\equiv(y-1=0, x-1=0)\equiv(1, 1)$ & its focus at $S\equiv\left(x-1=-\frac{1}{4}, y-1=0\right)\equiv(\frac{3}{4}, 1)$ 
Find out the points of intersection of parabola with the x-axis by setting $y=0$ in the equation as $(0, 0)$ & similarly, the points of intersection of parabola with the y-axis by setting $x=0$ in the equation as $(0, 0)$ & $(0, 2)$
Now, the curve of the parabola can be easily plotted by locating 
Step 1: Specify the vertex V & the focus S & join them to get the axis of parabola (as shown by the dotted line in the figure below)
Step 2: Specify the points of intersection of the parabola with the co-ordinate axes & take the length of latus rectum $4A=1$
Step 3: Draw, though these points, both the arms of parabola opening in the negative x-direction. (As shown in the figure below manually drawn)    
 
