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If $A = (0, 2), B = (0, - 2)$ and $|AP + PB| = 6,$ then the locus of P is

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

$9 x^{2} +5 y^{2} - 45 = 0$

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

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

answer is B.

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Detailed Solution

$A (0, 2), B (0, - 2), |AP + PB| = 6 \Rightarrow \frac{4 x^{2}}{36 - 16} + \frac{4 y^{2}}{36} = 1 \Rightarrow \frac{x^{2}}{5} + \frac{y^{2}}{9} = 1 \Rightarrow 9 x^{2} + 5 y^{2} - 45 = 0$

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