Two points A and B have coordinates (1, 0) an | vedclass

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(B) Let the coordinates of $Q$ be $(x, y)$.Given $A = (1, 0)$ and $B = (-1, 0)$.The condition is $|AQ - BQ| = 1$,which implies $AQ - BQ = \pm 1$.Subst

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

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(B) Let the coordinates of $Q$ be $(x, y)$.Given $A = (1, 0)$ and $B = (-1, 0)$.The condition is $|AQ - BQ| = 1$,which implies $AQ - BQ = \pm 1$.Substituting the distance formula: $\sqrt{(x - 1)^2 + y^2} - \sqrt{(x + 1)^2 + y^2} = \pm 1$.Rearranging: $\sqrt{(x - 1)^2 + y^2} = \pm 1 + \sqrt{(x + 1)^2 + y^2}$.Squaring both sides: $(x - 1)^2 + y^2 = 1 + (x + 1)^2 + y^2 \pm 2\sqrt{(x + 1)^2 + y^2}$.$x^2 - 2x + 1 + y^2 = 1 + x^2 + 2x + 1 + y^2 \pm 2\sqrt{(x + 1)^2 + y^2}$.$-4x - 1 = \pm 2\sqrt{(x + 1)^2 + y^2}$.Squaring again: $(-4x - 1)^2 = 4((x + 1)^2 + y^2)$.$16x^2 + 8x + 1 = 4(x^2 + 2x + 1 + y^2)$.$16x^2 + 8x + 1 = 4x^2 + 8x + 4 + 4y^2$.$12x^2 - 4y^2 = 3$.

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:

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