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

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$2 x y + 16 - x^{2} = 0$

$2 x y - 16 + x^{2} = 0$

$2 x y - 16 - x^{2} = 0$

$2 x y + 16 + x^{2} = 0$

answer is B.

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Detailed Solution

Let $M = (0, h), N \equiv (0, h + 4)$

Equation of AM $\frac{x}{- 4} + \frac{y}{h}$ = 1 ... (1)
Equation of BNO $\frac{x}{4} + \frac{y}{h + 4}$ = 1 ... (2)

Eliminating h from (1) and (2), we get
$x^{2} + 2 x y - 16 = 0$

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