locus of point of intersection of the normals at the ends of the parallel chords

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another parabola Prove:--> Let the equations of normals to the parabola y² = 4ax at point P (at²1, 2at1) and Q (at²2, 2at2) are y + t1x = 2at1+ at³1 - - - - (i) and y + t2x = 2at2+ at³2 - - - - (ii) Let (h, k) be the point of intersection of two perpendicular normals. so, k + t1h = 2at1+ at³1 - - - - (iii) and k + t2h = 2at2+ at³2 - - - - (iv) solving simultaneously (iii) & (iv) for intersection point (h,k) we get h = 2a + a(t²1+ t1t2+ t²2) and k = t1(1 - t²2) using t1t2= -1, {as the normals meet at right angles} => h/a = 1 + t²1+ t²2and k/a = ( t1+ t2) now, using, ( t1+ t2)² = t²1+ t²2+ 2 t1t2 => ( t1+ t2)² = t²1+ t²2- 2 => ( t1+ t2)² = (t²1+ t²2+ 1) - 3 => (k/a)² = (h/a) - 3 => k² = ah - 3a² - - - - {multiply throughout by a²} => y² = ax - 3a² => y² = a(x - 3a) ? (Proved)

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