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Distance Formula to Find Condition for Co-Linearity

Two points A ...

Question

Two points A and B have coordinates (1, 0) and (-1, 0) respectively and Q is a point which satisfies the relation

AQ - BQ = $\pm$ 1. The locus of Q is

$12 x^{2} + 4 y^{2} = 3$

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$12 x^{2} - 4 y^{2} = 3$

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$12 x^{2} - 4 y^{2} + 3$ = 0

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$12 x^{2} + 4 y^{2} + 3$ = 0

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Solution

The correct option is B
$12 x^{2} - 4 y^{2} = 3$

According to the given condition

$\sqrt{(x - 1)^{2} + y^{2}} - \sqrt{(x + 1)^{2} + y^{2}}$ = $\pm$ 1

On squaring both sides, we get

$2 x^{2} + 2 y^{2} + 1 = 2 \sqrt{(x - 1)^{2} + y^{2}} \sqrt{(x + 1)^{2} + y^{2}}$

Again on squaring, we get $12 x^{2} - 4 y^{2} = 3$ .

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Similar questions

Two points A and B have coordinates (1, 0) and (-1, 0) respectively and Q is a point which satisfies the relation

AQ - BQ = $\pm$ 1. The locus of Q is

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