Question

The coordinates of the point on the parabola y² = 8x, which is at minimum distance from the circle x² + (y + 6)² = 1 are-

• A. (2, –4)
• B. (18, –12)
• C. (2, 4)
• D. ) None of these

Answer 1

Answer: (A)

(2, –4)

Answer 2

A point on the parabola is at a minimum distance from the circle if and only if it is at a minimum distance from the centre of the circle. Any point on the parabola

y² = 8x is of the form P(2t², 4t). The centre of the circle x² + (y + 6)² = 1 is O (0, –6)

OP² = 4t⁴ + (–6 – 4t)² = 4 (t⁴ + 4t² + 12t + 9)

Let A = t⁴ + 4t² + 12t + 9

$\frac { \mathrm { dA } } { \mathrm { dt } }$= 4t³ + 8t + 12 = 4 (t³ + 2t + 3) = 4(t + 1) (t² – t + 3)

So $\frac { \mathrm { dA } } { \mathrm { dt } }$ = 0 if t = –1. Moreover,

$\left. \frac { \mathrm { d } ^ { 2 } \mathrm {~A} } { \mathrm { dt } ^ { 2 } } \right| _ { \mathrm { t } = - 1 }$ = 4 (3 (–1)² + 2) > 0. Hence required point is P

(2, –4).