Question

A ŗ (–4, 0), B ŗ (4, 0), M and N are variable points on y-axis such that M lies below N and MN = 4. If the line joining AM and BN intersect at P, then locus of P is-

• A. 2xy + 16 + x2 = 0
• B. 2xy – 16 + x2 = 0
• C. 2xy – 16 – x2 = 0
• D. 2xy + 16 – x2 = 0

Answer 1

Answer: (B)

2xy – 16 + x2 = 0

Answer 2

Let M = (0, h), N ŗ (0, h + 4)

Equation of AM

$\frac{x}{–4} + \frac{y}{h}$ = 1 .....(1)

Equation of BN0

$\frac{x}{4} + \frac{y}{h + 4}$ = 1 ....(2)

Eliminating h from (1) and (2), we get

x2 + 2xy – 16 = 0