Question

A, B are variable points lying on the lines y = 2x and y = x respectively such that AB = 4. The locus of the mid point of AB is –

• A. x² + y² ± 2x + 7 = 0
• B. Circle
• C. x² + 13 y – 25 = 0
• D. 25x² + 13y² – 36xy – 4 = 0

Answer 1

Answer: (D)

25x² + 13y² – 36xy – 4 = 0

Answer 2

Let A ŗ (a, 2a) and B ŗ (b, b)

Now, we have

AB = 4

i.e. (a – b)² + (2a – b)² = 16 … (1)

If M(h, k) be he mid-point of AB, then

2h = a + b ... (2)

and 2k = 2a + b … (3)

Solving equation (2) and (3), we have

a = 2(k – h) and b = 2(2h – k)

Now putting the above values in equation (1), we have

(4k – 6h)² + (6k – 8h)² = 16

i.e.100h² + 52k² – 144hk – 16 = 0

i.e.25h² + 13k² – 36hk – 4 = 0

Hence, the locus of M is

25x² + 13y² – 36xy – 4 = 0 .