Finding Locus - Parabola

Question

In the point A on the parabola:

$y^{2}=2px$

a tangent line to the parabola is being drawn such that it meets the y-axis at the point B. Find the locus of the meeting points C of the line from B which is parallel to the x-axis with the line from A which is parallel to the y-axis.

What I did so far, I defined the points:

$A(x_{0},y_{0})$ and $B(x_{1},y_{1})$.

The equation of the tangent line to the parabola is:

$yy_{0}=px+px_{0}$

I then used $x=0$ to find that the point C that represent the locus is:

$C(x_{A},\frac{px_{A}}{y_{A}})$

I know that I need somehow to use the parabola equation in order to find the answer, which should be the parabola $y^{2}=\frac{px}{2}$. I am not sure how to proceed.

Any help will be mostly appreciated.

For your convenience I used GeoGebra to draw this scenario for p=2.

I need to find the equation of the red parabola given the blue.

Answer 1

You know $$x_C=x_A$$ and that $$y_C=\frac{px_A}{y_A}$$ which means $$y_C=\frac{px_C}{\pm\sqrt{2px_C}}$$ which means $$y_C^{2}=\frac{px_C}{2}$$

Answer 2

The general equation of a tangent to a parabola of the form $y^2=4ax$ is $ty=x+at^2$ at the point $(at^2,2at)$ . This can be proved by differentiating the equation of the parabola. I'm going to define $a=\frac{p}{2}$

Hence, point $B=(0,at)$. Hence the line parallel to x-axis passing through $B$ is $y=at$. And the line parallel to y axis is $x=at^2$. The point of intersection is $(at^2,at)$. And the locus is $y^2=ax$ or $y^2=\frac{px}{2}$.