lf $A = (4, 0)$ , $B = (- 4, 0)$ are any two points and $| P A - P B | = 4$ . The locus of $P$ is:

3 x^{2} + y^{2} = 12

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

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3 x^{2} - 3 y^{2} = 9

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3 x^{2} + y^{2} = 8

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Solution

The correct option is C $3 x^{2} - y^{2} = 12$
$| P A - P B | = 4$
squaring both sides we get,
$(P A)^{2} + (P B)^{2} = 16 + 2 (P A) (P B)$
$(h - 4)^{2} + k^{2} + (h + 4)^{2} + k^{2} = 16 + 2 \sqrt{(h^{2} + 16 - 8 h + k^{2}) (k^{2} + h^{2} + 16 + 8 h)}$
$(h^{2} + k^{2} + 8)^{2}$
$= h^{2} k^{2} + h^{4} + 16 h^{2} + 8 h^{3} + 16 k^{2} + 16 h^{2} + 256 + 128 h - 8 h k^{2} - 8 h^{3} - 128 h - 64 h^{2} + k^{4} + k^{2} h^{2} + 16 k^{2} + 8 h k^{2}$
$3 h^{2} - k^{2} = 12$
$3 x^{2} - y^{2} = 12$

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