$\textbf{Exercise:}$
If $\overline{PQ}$ is a focal chord of the parabola $x^2=4py$ and the coordinates of $P$ are $(x_0,y_0)$, show that the coordinates of $Q$ are $$ \left(\frac{-4p^2}{x_0},\frac{p^2}{y_0}\right) $$ (original image)
$\textbf{Remainder:}$
A focal chord is the line segment from one point in the parabola to another one in the parabola passing through the focus.
$\textbf{My try:}$
My idea was to use the coordinates of the point $P=(x_0,y_0)$ and the focus $(0,p)$ to determine the equation of that line. Then I would proceed by considering the solutions to that equation and $x^2=4py$ to get the points of intersection. So let's calculate:
$$\textbf{Important things to note: }\boxed{y=mx+p}\boxed{\color{blue}{\text{distance in blue}=-x_0}}\boxed{\color{red}{\text{distance in red}=p-y_0}}$$
$$m=\frac{p-y_0}{-x_0}\implies \boxed{y=(\frac{y_0-p}{x_0})x+p}$$ Since we're looking for points of intersection and since the equation of the parabola can be written as $y=x^2/4p$ then: $$x^2/4p=(\frac{y_0-p}{x_0})x+p$$ Using the quadratic equation after simplifications we will get $$x=\frac{(4py_0-4p^2)\pm\sqrt{(4p^2-4py_0)-4x_0(-4p^2)}}{2x_0}$$ but that's not the result I'm looking for.
May someone show me where I got wrong in my reasoning, thanks!
The focus of $x^2=4py$ is at $(0,p)$. Therefore, the focal line with slope $m$ is $$ y=mx+p\tag{1} $$ $x$ for the two points of intersection with the parabola satisfy $$ x^2=4p(mx+p)\implies x^2-4pmx-4p^2=0\tag{2} $$ Using Vieta's Formulas, we get that the product of the two roots of $(2)$ is $$ x_0x_1=-4p^2\tag{3} $$ Furthermore, $$ x_0^2x_1^2=16p^2y_0y_1\tag{4} $$ Thus, the product of the two $y$ values is $$ y_0y_1=\frac{x_0^2x_1^2}{16p^2}=\frac{\left(-4p^2\right)^2}{16p^2}=p^2\tag{5} $$ Therefore, if $(x_0,y_0)$ is one end of the focal chord, the other end would be $$ (x_1,y_1)=\left(\frac{-4p^2}{x_0},\frac{p^2}{y_0}\right)\tag{6} $$