Question

Slope of the normal chord of y² = 8x that gets bisected at (8, 2) is 

(a)  1

(b)  – 1

(c)  2

(d)  – 2

Answer 1

Correct option (c) 2

Explanation:

Let AB be the normal chord where A = (2t₁² , 4t₁), B = (2t₂² , 4t₂). 

It’s slope = 2/t₁ + t₂ 

We also have t₂ = – t₁ – 2/t₁  and 16 = 2(t₁ ² + t₂²), 4 = 4(t₁ + t₂)

⇒ t₁ + t₂ = 1.

Thus slop is 2