Suppose we have the following functions with shared parameter $t$:
$$x(t) = -\sin(t)+\cos(t)+1$$ $$y(t) = -\sin(t)\cos(t)$$
When we plot them together as a planar curve we can see what appears to be a parabola (red). Visually guessing and tinkering with transformations of a parabola I get a curve that appears to fit nicely (black).
The specific transformed parabola is
$$y = \frac{1}{2} \left[ \left(x - 1 \right)^2 -1 \right].$$
In terms of my parametric functions, does $$y(t) = \frac{1}{2} \left[ \left(x(t) - 1 \right)^2 -1 \right]?$$
Yes, the equality holds. Let us start with the RHS and simplify down to the LHS.
$\begin{align} \frac{1}{2} \left[ \left(x(t) - 1 \right)^2 -1 \right] &= \frac{1}{2} \left[ \left(-\sin(t)+\cos(t)+1 - 1 \right)^2 -1 \right]\\ &= \frac{1}{2} \left[ \left(-\sin(t)+\cos(t) \right)^2 -1 \right]\\ &= \frac{1}{2} \left[ \sin^2(t) - 2\sin(t)\cos(t)+\cos^2(t) -1 \right] \\ &= \frac{1}{2} \left[ 2\sin(t)\cos(t) +1 -1 \right] \\ &= \frac{1}{2} \left[ 2\sin(t)\cos(t) \right] \\ &= \sin(t)\cos(t) \\ &= y(t)\end{align}$