Lines intersect at point $P$, draw perpendiculars to the tangents through points $A$ and $B$, and their intersection is point $Q$. Find the expression for the coordinates of point $Q$ in terms of the coordinates of point $P$, and answer: (1) When the y-coordinate of point $P$ remains constant and the x-coordinate changes, what is the trajectory of point $Q$? (2) When the x-coordinate of point $P$ remains constant and the y-coordinate changes, what is the trajectory of point $Q$?
Let the coordinates of points be $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right), P\left(x_{P}, y_{P}\right), Q\left(x_{Q}, y_{Q}\right)$, then the tangent line of the parabola at point $A$ is $y_{A} y=x+x_{A}$ (1) passing through point $P$, so $y_{A} y_{P}=x_{P}+x_{A}$. (2) Similarly, $y_{B} y_{P}=x_{P}+x_{B}$. (3)

From (2) and (3), we get $x_{P}=\frac{x_{A} y_{B}-y_{A} x_{B}}{y_{A}-y_{B}}=\frac{1}{2} y_{A} y_{B}$, (4) $y_{P}=\frac{x_{A}-x_{B}}{y_{A}-y_{B}}=\frac{1}{2}\left(y_{A}+ y_{B}\right)$, (5) The equation of the normal line $Q A$ is $y-y_{A}+y_{A}\left(x-x_{A}\right)=0$, (6) so $y_{Q}-y_{A}+ y_{A}\left(x_{Q}-x_{A}\right)=0$.

Similarly, $y_{Q}-y_{B}+y_{B}\left(x_{Q}-x_{B}\right)=0$. (8) From (7) and (8), we get $x_{Q}=\frac{1}{2}\left(y_{A}^{2}+y_{A} y_{B}+ y_{B}^{2}\right)+1$, (9) $y_{Q}=-\frac{1}{2} y_{A} y_{B}\left(y_{A}+y_{B}\right)$, (10) From (4), (5), (9), and (10), we get $x_{Q}=2 y_{P}^{2}- x_{P}+1$ (11) $y_{Q}=-2 x_{P} y_{P}$. (12) When the y-coordinate of point $P$, $y_{P}$, is constant, by eliminating $x_{P}$ from (11) and (12), we get $y_{Q}=2 y_{P} x_{Q}-2 y_{P}\left(2 y_{P}^{2}+ 1\right)$, at this time the trajectory of point $Q$ is a straight line; (2) When the x-coordinate of point $P$, $x_{P}$, is constant, by eliminating $y_{P}$ from (11) and (12), we get $y_{Q}^{2}=2 x_{P}^{2}\left(x_{Q}+x_{P}-1\right)$, at this time the trajectory of point $Q$ is a parabola.