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A(4, 0) B(–4, 0) are two points. The locus of P which moves such that $| PA - PB | = 6$ is

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$9 x^{2} - 7 y^{2} + 63 = 0$

$9 x^{2} + 7 y^{2} - 63 = 0$

$9 x^{2} + 7 y^{2} + 63 = 0$

$7 x^{2} - 9 y^{2} - 63 = 0$

answer is A.

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Detailed Solution

$P A = P B \pm 6 or \frac{4 (x - a)^{2}}{k^{2} - 4 b^{2}} + \frac{4 y^{2}}{k^{2}} = 1$

$where k = 6; a = 0; b = 4$

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