Two points A and B have coordinates 1-0 and -1-0 respectively and Q is a point which satisfies the relation AQ - BQ - 1- The locus of Q isA- 12 x2-4 y2-3B- 12 x2-4 y2-3-0C- 12 x2-4 y2-3-0D- 12 x2-4 y2-3

The correct option is D According to the given condition √((x 1)^2 + y^2) √((x + 1)^2 + y^2) = ±1 On squaring both sides, we get 2x^2 + 2y^2 + 1 = 2√((x ...

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Standard XII

Mathematics

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

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Solution

The correct option is B

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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Two points A and B have coordinates (1, 0) and (-1, 0) respectively and Q is a point which satisfies the relation AQ - BQ = ± 1. The locus of Q is

The correct option is B According to the given condition √(x−1)2+y2−√(x+1)2+y2 = ±1 On squaring both sides, we get 2x2+2y2+1=2√(x−1)2+y2√(x+1)2+y2 Again on squaring, we get 12x2−4y2=3.

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

The correct option is B

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