<prompt> Lengo moja la duaradufu fracx22 y2 1 iko F 1,0. Kuna uhakika P p,0, ambapo p 0, kiasi kwamba kwa chord yoyote overlineAB ambayo inapitia F, angles angle APF na angle BPF ni sawa. Tafuta p.asyukubwa wa kitengo 2 cm;jozi A, B, F, P;path ell xscalesqrt2Circle0,0,1;F 1,0;A sqrt2Cos80,Sin80;B sehemu ya makutanointerpA,F,0.1--interpA,F,5,ell;P 2,0;kuteka ell;choraA--B;choraA--P--B;choraF--P;nuktaA, A, N;nukta B, B, SE;nukta F, F, SW;nukta P, P, E;asy  <response>: Kwanza, tunazingatia mstari fulani, y x - 1, ambao hupitia F. Kubadilisha, tunapatafracx22 x - 12 1.Hii hurahisisha hadi 3x2 - 4x x3x - 4 0, hivyo x 0 au x frac43. Kwa hivyo, tunaweza kuruhusu A left frac43, frac13 kulia na B 0,-1.Kisha mteremko wa mstari AP ni frac1343 - p frac14 - 3p, na mteremko wa mstari BP ni frac -1-p frac1p. Kwa kuwa angle APF angle BPF, miteremko hii ni hasi ya kila mmoja, kwa hivyofrac13p - 4 frac1p.Kisha p 3p - 4, hivyo p 2.Kwa suluhu kamili, tunathibitisha kwamba hii inafanya kazi kwa chords zote overlineAB ambazo hupitia F. Let A x_a,y_a na B x_b,y_b. Kisha condition angle APF pembe BPF ni sawa nafracy_ax_a - 2 fracy_bx_b - 2 0,au y_a x_b - 2 y_b x_a - 2 0. Kisha y_a x_b - 2y_a y_b x_a - 2y_b 0.Acha y mx - 1 iwe mlinganyo wa mstari AB. Kubadilisha, tunapatafracx22 m2 x - 12 1.Hii hurahisisha hadi 2m2 1 x2 - 4m2 x 2m 2 - 2 0. Kwa fomula za Vieta,x_a x_b frac4m22m2 1 quad textna quad x_a x_b frac2m2 - 22m2 1. Kishaanzaaligny_a x_b - 2y_a y_b x_a - 2y_b mx_a - 1 x_b - 2mx_a - 1 mx_b - 1 x_a - 2mx_b - 1 2mx_a x_b - 3m x_a x_b 4m 2m cdot frac2m2 - 22m2 1 - 3m cdot frac4m22m2 1 4m 0.endalignKwa hivyo, angle APF pembe BPF kwa chords zote overlineAB ambazo hupitia F. 