Question

Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ of the parabola be drawn through it, then
$\frac{1}{P{K}^2}+\frac{1}{Q{K}^2}$
is the same for all positions of the chord.

Answer 1

let the point $K$ be $(d,0)$

the problem have to be solved using the equation of line in polar form that is

$\frac{x-d}{\cos \alpha }=\frac{y}{\sin \alpha }=r$

therefore the coordinates of $P$ and $Q$ are $(c+KP\cos \theta ,KP\sin \theta )$ and $(c-KQ\cos \theta ,-KQ\sin \theta )$

point as $P$ and $Q$ lie in the parabola then

$(KP{)}^2{\sin }^2\theta =4a(c+KP\cos \theta )$ and $(KP{)}^2{\sin }^2\theta =4a(c+KP\cos \theta )$

then solving the quadratic equations and taking only the positive roots(because the lengths are always positve) we get

$KP=\frac{4a+{(16a^2{\cos }^2\theta +16ac{\sin }^2\theta )}^{1/2}}{2a{\sin }^2\theta }$ and $KQ=\frac{-4a+{(16a^2{\cos }^2\theta +16ac{\sin }^2\theta )}^{1/2}}{2a{\sin }^2\theta }$

therefore $\frac{1}{K{P}^2}+\frac{1}{K{Q}^2}=\frac{2a{\cos }^2\theta +c{\sin }^2\theta }{2a{c}^2}$

for $c=2a\frac{1}{K{P}^2}+\frac{1}{K{Q}^2}=\frac{1}{c^2}$ which is independent of theta

hence proved