# The focal chord that cuts the parabola $x^2 = -6y$ at $(6, -6)$ cuts the parabola again at $X$

The focal chord that cuts the parabola $x^2 = -6y$ at $(6, -6)$ cuts the parabola again at $X$. Find the coordinates of $X$. I have been going insane someone please help me :(

Compare the equation with $x^2 = -4ay$ and find the coordinates of the focus.
You could find the coordinates of the focus and write the equation of the parabola in parametric form $(2at, -at^2)$.
After that, write the equation of a general focal chord joining the focus $(0, -a)$ and the point on the parabola $(2at, -at^2)$. Plug in $(6,-6)$ into the equation and then solve for the value(s) of the parameter $t$. To get the point, substitute $t$ into $(2at, -at^2)$.