Chord of a parabola $y^{2}= 4ax$

Question

Prove that on the axis of any parabola $y^2=4ax$ there is a certain point $K$ which has the property that,if a chord $PQ$ of the parabola be drawn through it ,then $$\frac{1}{PK^2}+\frac{1}{QK^2}$$ is same for all positions of the chord.Find also the coordinates of the point $K$.

We can apply the parametric equations of a parabola Let the points $P$ and $Q$ be $(at_1^{2},2at_1)$ and $(at_2^{2}, 2at_2)$

So the equation of the chord would be $$y(t_1+t_2)=2x+2at_1t_2$$

Hence from there we have that the coordinates of $K$ are $(−at_1t_2,0)$

Now our aim is to show that $\frac{1}{PK^2}+\frac{1}{QK^2}$ is independent of $t_1$ and $t_2$. I tried and applied the distance formula but no benefit.

Answer 1

Hint: This is beyond easy ! :-$)$ The sum in question stays constant, regardless of the position of P, correct ? So just let $K=(b,0),$ and then take two positions for P: when P is right above $($or right below$)$ K, and when $P\to\infty$. Where is Q in both these cases ? Can you deduce the value of b from equating the two sums ? I just did, and used GeoGebra to verify the result.