The locus of a point whose difference of distances from points $(\pm 3, 0)$ is $4,$ is :

x^{2} / 4 - y^{2} / 5 = 1

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x^{2} / 5 - y^{2} / 4 = 1

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

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

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Solution

The correct option is A $x^{2} / 4 - y^{2} / 5 = 1$
Let $p (3, o) p^{'} (- 3, o) 0 (x, y)$

$| o p - o p^{'} | = 4$

$o p^{2} + o p^{' 2} - 2. o p . o p^{'} = 16$

$(x - 3)^{2} + (y)^{2} + (x + 3)^{2} + y^{2} - 2 \sqrt{[(x - 3)^{2} + y^{2}] [(x + 3)^{2} + y^{2}]} = 16$

$2 y^{2} + 2 (x^{2} + 9) - 2 \sqrt{(x^{2} - 9)^{2} + y^{2} [2 (x^{2} + 9)] + y^{4}} = 16$

$2 \sqrt{(x^{2} - 9)^{2} + y^{2} (2 (x^{2} + 9)) + y^{4}} = 16 - 18 - 2 (x^{2} + y^{2})$

$\sqrt{(x^{2} - 9)^{2} + y^{2} [2 (x^{2} + 9)] + y^{4}} = - [2 (x^{2} + y^{2} + 2)]$

$(x^{2} - 9^{2}) + 2 y^{2} (x^{2} + 9) + y^{4} = (x^{2} + y^{2}) + 1 + 2 (x^{2} + y^{2})$

$x^{4} - 18 x^{2} + 81 + 2 x^{2} y^{2} + 18 y^{2} + y^{4} = x^{4} + y^{4} + 2 x^{2} + y^{2} + 1 + 2 x^{2} + 2 y^{2}$

$18 y^{2} - 2 y^{2} - 18 x^{2} - 2 x^{2} = 1 - 81$

$20 x^{2} - 16 y^{2} = 80$

$\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$

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