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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. |
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Answer» 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)`. |
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