The locus of the point of intersection of the | vedclass

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(B) Given lines are $(\sqrt{3})kx + ky = 4\sqrt{3}$ and $\sqrt{3}x - y = 4\sqrt{3}k$.From the first equation,$k = \frac{4\sqrt{3}}{\sqrt{3}x + y}$.Fro

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$2$

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(B) Given lines are $(\sqrt{3})kx + ky = 4\sqrt{3}$ and $\sqrt{3}x - y = 4\sqrt{3}k$.From the first equation,$k = \frac{4\sqrt{3}}{\sqrt{3}x + y}$.From the second equation,$k = \frac{\sqrt{3}x - y}{4\sqrt{3}}$.Equating the two expressions for $k$:$\frac{4\sqrt{3}}{\sqrt{3}x + y} = \frac{\sqrt{3}x - y}{4\sqrt{3}}$$(\sqrt{3}x - y)(\sqrt{3}x + y) = 16(3) = 48$$3x^2 - y^2 = 48$Dividing by $48$,we get $\frac{x^2}{16} - \frac{y^2}{48} = 1$.This is the equation of a hyperbola with $a^2 = 16$ and $b^2 = 48$.The eccentricity $e$ is given by $b^2 = a^2(e^2 - 1)$.$48 = 16(e^2 - 1)$$3 = e^2 - 1$$e^2 = 4$$e = 2$.

The locus of the point of intersection of the lines $(\sqrt{3})kx + ky - 4\sqrt{3} = 0$ and $\sqrt{3}x - y - 4\sqrt{3}k = 0$ is a conic,whose eccentricity is .............

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