Let $t$ be TNYWR.

The line with equation $y=-2 x+t$ and the parabola with equation $y=(x-1)^{2}+1$ intersect at point $P$ in the first quadrant. What is the $y$-coordinate of $P$ ?
To find the points of intersection of the line with equation $y=-2 x+t$ and the parabola with equation $y=(x-1)^{2}+1$, we equate values of $y$, to obtain

$$
\begin{aligned}
(x-1)^{2}+1 & =-2 x+t \\
x^{2}-2 x+1+1 & =-2 x+t \\
x^{2} & =t-2
\end{aligned}
$$

The points of intersection thus have $x$-coordinates $x=\sqrt{t-2}$ and $x=-\sqrt{t-2}$.

The point $P$ in the first quadrant has a positive $x$-coordinate, and so its $x$-coordinate is $\sqrt{t-2}$.

Thus, the $y$-coordinate of $P$ is $y=-2 \sqrt{t-2}+t$.

Since the answer to (a) is 11 , then $t=11$.

This means that the $y$-coordinate of $P$ is $y=-2 \sqrt{11-2}+11=-2 \cdot 3+11=5$.