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If $A (4, 0) and B (- 4, 0)$ are two points, then the locus of a point P such that $| PA - PB | = 4$

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

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

$4 (x^{2} - 3 y^{2}) = 1$

$3 x^{2} - y^{2} = 1$

answer is A.

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Detailed Solution

$| PA - PB | = 4$

$P A = P B \pm 4$

S.O.B.S

${(P A)}^{2} = {(P B)}^{2} + 16 \pm 8 P B$

${(P A)}^{2} - {(P B)}^{2} = 16 \pm 8 P B$

$(x - 4)^{2} + y^{2} - (x + 4)^{2} + y = 16 \pm 8 P B$

simplifying will get

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

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