If the normal at the point ‘t1‘ on the parabola y2 = 4ax meets the

1.	If the normal at the point ‘t1‘ on the parabola y2 = 4ax meets the parabola again at the point ‘t2‘ , then prove that t2 = (t1 + (2/t1))
Answer» Equation of normal to y² = 4 at’ t’ is y + xt = 2at + at³. So equation of normal at ‘t₁’ is y + xt₁ = 2at₁ + at₁³ The normal meets the parabola y²= 4ax at ‘t₂’ (ie.,) at (at₂², 2at₂) ⇒ 2at₂ + at₁t₂² = 2at₁ + at₁³ So 2a(t₂ – t₁) = at₁³ – at₁t₂² ⇒ 2a(t₂ – t₁) = at₁(t₁² – t₂²) ⇒ 2(t₂ – t₁) = t₁(t₁ + t₂)(t₁– t₂) ⇒ 2 = -t₁(t₁ + t₂) ⇒ t₁ + t₂ = -2/t₁ ⇒ t₂ = -t₁ - (2/t₁) = -(t₁ + (2/t₁))

If the normal at the point ‘t1‘ on the parabola y2 = 4ax meets the parabola again at the point ‘t2‘ , then prove that t2 = (t1 + (2/t1))

Answer»

Equation of normal to y² = 4 at’ t’ is y + xt = 2at + at³.

So equation of normal at ‘t₁’ is y + xt₁ = 2at₁ + at₁³

The normal meets the parabola y²= 4ax at ‘t₂’

(ie.,) at (at₂², 2at₂)

⇒ 2at₂ + at₁t₂² = 2at₁ + at₁³

So 2a(t₂ – t₁) = at₁³ – at₁t₂²

⇒ 2a(t₂ – t₁) = at₁(t₁² – t₂²)

⇒ 2(t₂ – t₁) = t₁(t₁ + t₂)(t₁– t₂)

⇒ 2 = -t₁(t₁ + t₂)

⇒ t₁ + t₂ = -2/t₁

⇒ t₂ = -t₁ - (2/t₁) = -(t₁ + (2/t₁))