Question

A = (-4, 0) and B = (4, 0). The points M and N are variable points on the y-axis such that N lies above M and MN = 4. The lines AM and BN intersect at P. Then, the locus of P is

(a)  x² + 2xy - 16 = 0

(b)  x² = 2xy + 16

(c)  x² + 16 = 2xy

(d)  x² + 2xy + 16  = 0

Answer 1

Correct option (a)  x² + 2xy - 16 = 0

Explanation :

See Fig. We have A = (−4, 0) and B = (4, 0) so that y-axis is the perpendicular bisector of AB. Since MN = 4, we can take N = (0, a + 4) and M = (0, a). Equation of the line AM is

x/-4 + y/a = 1   .....(1)

 and the equation of BN is

x/4 + y/a + 4 = 1   .....(2)

From Eq. (1), we have

 a = 4y/4 + x  ....(3)

From Eq. (2), we get

a + 4 = 4y/4 - x  ....(4)

 Therefore, from Eqs. (3) and (4), we have