Question

Three normals to y² = 4x pass through the point (15,12). One of the normals is

(A) x+y = 27

(b) x+4y = 63

(c) 3x–y = 33

(d) y+3x = 51

Answer 1

Correct option is (c) 3x–y = 33

Let equation of normal be y = mx–2am–am³ .     

a = 1

∴ y = mx–2m–m³

passes through (15,12)

12 = 15m – 2m–m³

m³ –13m +12 = 0

(m–1) (m–3) (m+4) = 0

m = 1, 3, –4.

m =1 ⇒ y = x–3

m =3 ⇒ y = 3x–33

m =–4 ⇒ y + 4x = 72