$PN$ is the ordinate of a point $P$ on the parabola $y^2=4ax$;
$NP$ is produced to $Q$ so that $PQ = SP$ where $S$ is the focus. Find the locus of $Q$.
The answer is $y^2 + x^2 - 2xy - 2ay - 2ax + a^2 = 0$.
Can anyone show me how to solve this question? I can't cancel out the parameter of the equation... Thank you very much
$y^2=4ax$ equation of parabola
$P\left(\dfrac{t^2}{4a},t\right)$ generic point
$S(a,0)$ focus
$h:x=-a$ directrix
$QP=PS=PH=t+a$ by definition of parabola
$QN=PN+QP= \dfrac{t^2}{4a}+(t+a)$
$$Q\left(x= \frac{t^2}{4 a},y= \frac{t^2}{4 a}+a+t\right)$$
$$y-x=a+t\to t=y-x-a\text{ so }y=\frac{(-a-x+y)^2}{4 a}-x+y$$
$$-(-a-x+y)^2-4 a (y-x)+4 a y=0 \to \color{red}{x^2-2 x y+y^2-2 a x-2 a y+a^2=0}$$ Which is the equation of the rotated parabola that can be seen in the graph below
Hope this can be useful $$...$$