Tangent to the curve y = x2 + 6 at a pointegral P(1, 7) touc... | MATH - PARABOLA | SolveeIt | Solveeit

Today, August 19

Question

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

A

(–6, –11)

B

(–9, –13)

C

(–10, –15)

D

(–6, –7)

Answer

(–6, –7)

Explanation

Solution

Equation of tangent at P to the parabola

2y = 2x² + 6 × 2

y + 7 = 2x . 1 + 12

2x – y + 5 = 0 …(1)

Equation of normal at Q

x + 2y + l = 0

– 8 – 12 + l = 0 Ž l = 20

x + 2y + 20 = 0 …(2)

Solving (1) and (2)

x = – 6, y = – 7