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$A (0, a e) B (0, - a e)$ are two points. The equation to the locus of p such that $P A - P B = 2 a$ is

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$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 + e^{2})} = 1$

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2} (1 + e^{2})} = 1$

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2} (1 - e^{2})} = 1$

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 - e^{2})} = 1$

answer is A.

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Detailed Solution

$P A^{2} = {(2 a + P B)}^{2} P A^{2} = {(x)}^{2} + {(y - a e)}^{2} P B^{2} = {(x)}^{2} + {(y + a e)}^{2} (o r) \frac{4 x^{2}}{k^{2}} + \frac{4 {(y - b)}^{2}}{k^{2} - 4 a^{2}} = 1$

$(k = 2 a, a = a e, b = 0)$

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