Question

Tangent to the curve y = x² + 6 at a point P(1, 7) touches the curve x² + y² + 16x + 12y + c = 0 at a point Q. Then the co-ordinates of Q are

(a) (– 9, – 13)

(b) (– 10, – 15)

(c) (– 6, – 7)

(d) (6, – 7)

Answer 1

Correct option (C) (-6, – 7)

Explanation:

Given parabola is y = x² + 6

⇒ dy/dx = 2x

Slope of the tangent = m = 2 at (1, 7)

Equation of the tangent at P(1, 7) is y – 7 = 2(x – 1)

⇒ 2x – y + 5 = 0 ...(i)

Given circle is x² + y² + 16x + 12y + c = 0

(x + 8)² + (y + 6)² = r²

Here, CQ is perpendicular to the tangent.

Thus, slope of CQ = – 1/2

Equation of CQ is y + 6 = – 1/2(x + 8)

⇒ x + 2y = –20 ...(ii)

Solving (i) and (ii), we get, x = –6, y = –7

Hence, the point Q is (– 6, –7)