Question

A line of fixed lingth2 units moves so that its ends are on the positive x-axis and that part of the line x + y = 0 which lies in the second quadrant. Then the locus of the mid-point of the line has the equation-

• A. x² + 5y² + 4xy – 1 = 0
• B. x² + 5y² + 4xy + 1 = 0
• C. x² + 5y² – 4xy – 1 = 0
• D. 4x² + 5y² + 4xy + 1 = 0

Answer 1

Answer: (A)

x² + 5y² + 4xy – 1 = 0

Answer 2

If ŠBAO = q then BM = 2 sin q and

MO = BM = 2 sin q, MA = 2 cos q

Hence A = (2 cos q – 2 sin q, 0) and

B = (–2 sin q, 2 sin q)

Since P(x, y) is the mid point of AB,

2x = (2 cos q) + (–4 sin q) or cos q – 2 sin q = x

2y = (2 sin q) or sin q = y

Eliminating q, we have (x + 2y)² + y² = 1 or ,

x² + 5y² + 4xy – 1 = 0.