Question

If the shortest distance of the parabola y² = 4x from the centre of the circle x² + y² - 4x - 16y + 64 = 0 is 'd', then find d²

(1) 10

(2) 20

(C) 30

(4) 15

Answer 1

Correct option is (2) 20

y² = 4x;

x² + y² - 4x - 16y + 64 = 0

circle (2, 8) radius = 2 

\(\because\) shortest distance will lie along common normal

\(\because\) normal at p(t) to parabola is

 y + tx = 2t + t², now it will pass through (2, 8)

⇒ 8 + 2t = 2t + t³

⇒ t = 2

\(\therefore\) p(t) = p(t², 2t) = P(4, 4)

\(\therefore\) d = \(\sqrt{4 + 16}\) = \(\sqrt{20}\)

\(\therefore\) d² = 20