Prove that for a suitable point `P` on the axis of the parabola, chord `A B` through the point `P` can be drawn such that `[(1/(A P^2))+(1/(B P^2))]` is same for all positions of the chord.

Explanation

Let the point P be (p,0) and the equation of the chord through P be
`(x-p)/(costheta)=(y-0)/(sintheta)=r(rinR)` (1)
Therefore, `(rcostheta+P,rsintheta)` lies on the parabola `y^(2)=4ax`.
So, `r^(2)sin^(2)theta-4arcostheta-4ap=0` (2) If `AP=r_(1)andBP=-r_(2)`, then `r_(1)andr_(2)` are the roots of (2).
Therefore,
`r_(1)+r_(2)=(4acostheta)/(sin^(2)theta),r_(1)r_(2)=(-4ap)/(sin^(2)theta)`
Now, `(1)/(AP^(2))+(1)/(BP^(2))=(1)/(r_(2)^(2))+(1)/(r_(2)^(2))`
`=((r_(1)+r_(2))^(2)-2r_(1)r_(2))/(r_(1)^(2)r_(2)^(2))`
`=(cos^(2)theta)/(p^(2))+(sin^(2)theta)/(2ap)`
Since `(1)/(AP^(2))+(1)/(BP^(2))` should be independent of `theta`, we take p2a. Then,
`(1)/(AP^(2))+(1)/(BP^(2))=(1)/(4a^(2))(cos^(2)theta+sin^(2)theta)=(1)/(4a^(2))`
Hence, `(1)/(AP^(2))+(1)/(BP^(2))` is independent of `theta` for the positions of the chord if `P-=(2a,0)`.