Problem involving chords at vertex of a standard parabola Chords AP and AQ are drawn through the vertex A of a parabola y² = 4ax 
 at right angles to one another. Prove that the line PQ cuts the axis in a fixed point. 
If I understand correctly, we're supposed to find the locus of the point on the axis and prove that it is constant. But I don't know how to start the deduction for the equation to PQ. 
 A: Hint
Let $y=mx$ be the equation of AP
$$(mx)^2=4ax$$
$x=0, \dfrac{4a}{m^2}\implies P(4a/m^2,4a/m)$
Similarly $AQ: y=-\dfrac xm,Q:?$
Now find the equation of $PQ$ and it's intersection with $y=0$ the axis of the parabola
A: Line 1
Y=mx
Line3
Y=-x/m
Because for lines to be perpendicular there product has to be -1
Now find where they cut the parabola 
(mx)^2=4ax
m^2x=4a
x=4a/m^2
Co ordinate 
(4a/m^2,4a/m) 
Similarly for line 2 
Co ordinate
(4am^2,-4am)
The minus of y co odrinate is intuitional 
Write the eqn of the line 
gradient=((4a/m)+4am)/((4a/m^2)-4am^2)
=((1/m)+m)/(((1/m)+m)((1/m)-m))
=m/1-m^2
Y+4am=(m/1-m^2)(x-4am^2)
Now you need to know family of lines 
to solve this 
Line a + e(line b)=0
Where e can take any real value 
And all line pass through a fixed point
The point where line a and line b meet
So to solve this group all m terms together
Now you will see y is with no m term so and now you can solve for the value of x
You should read a bit about family of lines if you didnt understand the solution 
