Q.2) Parabola.

Is the locus of point of intersection of two normals to a given parabola y^2 = 4ax which meet at right angles



given by y^2 = a (x - 3a) ?

SOLUTION : ►Yes, it is correct! Locus = y² = a(x - 3a)




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)