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$A (4, 0), B (- 4, 0)$ are two points. The locus of P which moves such that $|P A - P B| = 4$ is

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y+3=0

$3 x^{2} - y^{2} = 12$

y-5=0

$y^{2} - 3 = 0$

answer is B.

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Detailed Solution

$A (4, 0), B (- 4, 0) a n d | P A - P B | = 4, A B > k \frac{4 x^{2}}{16} + \frac{4 y^{2}}{16 - 64} = 1 \Rightarrow \frac{x^{2}}{4} - \frac{y^{2}}{12} = 1$
$\Rightarrow 3 x^{2} - y^{2} = 12$

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