The locus of the point of intersection of the | vedclass

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(B) Given lines are: $L_1: \sqrt{3}x - y = 4\sqrt{3}t$ $L_2: \sqrt{3}tx + ty = 4\sqrt{3} \implies t(\sqrt{3}x + y) = 4\sqrt{3} \implies t = \frac{4\sq

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$2$

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(B) Given lines are: $L_1: \sqrt{3}x - y = 4\sqrt{3}t$ $L_2: \sqrt{3}tx + ty = 4\sqrt{3} \implies t(\sqrt{3}x + y) = 4\sqrt{3} \implies t = \frac{4\sqrt{3}}{\sqrt{3}x + y}$ Substitute $t$ into $L_1$: $\sqrt{3}x - y = 4\sqrt{3} \left( \frac{4\sqrt{3}}{\sqrt{3}x + y} ight)$ $(\sqrt{3}x - y)(\sqrt{3}x + y) = 16 	imes 3$ $3x^2 - y^2 = 48$ Divide by $48$: $\frac{x^2}{16} - \frac{y^2}{48} = 1$ This is a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ where $a^2 = 16$ and $b^2 = 48$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$. $e = \sqrt{1 + \frac{48}{16}} = \sqrt{1 + 3} = \sqrt{4} = 2$.

The locus of the point of intersection of the lines $\sqrt{3}x - y - 4\sqrt{3}t = 0$ and $\sqrt{3}tx + ty - 4\sqrt{3} = 0$ (where $t$ is a parameter) is a hyperbola whose eccentricity is

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