Question

The normal at (16,16) to the parabola y² = 16x again meets at

• A. (36,-24)
• B. (36,24)
• C. (-36,24)
• D. (18,24)

Answer 1

Answer: (A)

(36,-24)

Answer 2

If the normal at t₁ cut the parabola again at

t₂, then t₂ = -t₁ - $\frac { 2 } { t _ { 1 } }$ .

Given 2at₁ = 16 ⇒ t₁ = 2

∴ t₂ = -2 - $\frac { 2 } { 2 } = - 3$

∴ Required point ( = (36, -24)