Question

P, Q are the points t₁ , t₂ on the parabola y² = 4ax. The normals at P, Q meet on the parabola. Show that the middle point of PQ lies on the parabola y² = 2a(x + 2a).

Answer 1

The points P and Q are (at₁ ² , 2at₁) and (at₂ ² , 2at₂)

As the normals at t₁& t₂ meet on the parabola, t₁ t₂ = 2 ... (i)

Also if (x₁ , y₁) be the midpoint of PQ, then

x₁ =  1/2  (at₁ ² + at₂ ²) and y₁ = 1/2  (2at₁ + 2at₂) ... (ii)

From (iii) we get (t₁ + t₂) ² = (y₁ /a)²

(y₁ /a)² = t₁ ² + t₂ ² + 2t₁ t₂ = (2x₁ /a) + 4, using (i) and (ii) 

⇒ y₁² = 2a (x₁ + 2a)

Hence the locus of (x₁ , y₁) is y² = 2a(x + 2a).