Problem involving chords at vertex of a standard parabola

Question

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.

Answer 1

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

Answer 2

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