Question

Three normals are drawn from the point (14, 7) to the curve y² – 16x – 8y = 0. Find the co-ordinates of the feet of the normals.

Answer 1

The given curve is y² – 16x – 8y = 0.

⇒ y² – 8y = 16x

⇒ (y – 4)² = 16(x + 1) ...(i)

Consider any point on (i) as P(4t² – 1, 8t + 4)

Now, (dy/dx)_P = 1/t

Thus, the equation of normal at point P is

y –(8t + 4) = – t(x –(4t² – 1))

which is passing through (14, 7)

So, 7 –(8t + 4) = – t(14 – (4t² – 1)

⇒ 3 – 8t = – 15t + 4t³

⇒ 4t³ – 7t – 3 = 0

⇒ 4t³ + 4t² – 4t² – 4t – 3t – 3 = 0

⇒ 4t²(t + 1) – 4t(t + 1) – 3(t + 1) = 0

⇒ (4t² – 4t – 3)(t + 1) = 0

⇒ (4t² – 6t + 2t – 3)(t + 1) = 0

⇒ {2t(2t – 3) + 1(2t – 3)} (t + 1) = 0 

⇒ (2t + 1)(2t – 3)(t + 1) = 0

⇒ t = –1, – 1/2 , 3/2

Hence, the foot of normals are

(3, – 4), (8, 16) and (0, 0).