The locus of the point of intersection of the | vedclass

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(D) Given lines are:$(i) \sqrt{3}x - y = 4\sqrt{3}k$$(ii) \sqrt{3}kx + ky = 4\sqrt{3}$From $(i)$,$k = \frac{\sqrt{3}x - y}{4\sqrt{3}}$.Substitute $k$

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$16$

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(D) Given lines are:$(i) \sqrt{3}x - y = 4\sqrt{3}k$$(ii) \sqrt{3}kx + ky = 4\sqrt{3}$From $(i)$,$k = \frac{\sqrt{3}x - y}{4\sqrt{3}}$.Substitute $k$ into $(ii)$:$\sqrt{3}x \left(\frac{\sqrt{3}x - y}{4\sqrt{3}}\right) + y \left(\frac{\sqrt{3}x - y}{4\sqrt{3}}\right) = 4\sqrt{3}$$\frac{3x^2 - \sqrt{3}xy + \sqrt{3}xy - y^2}{4\sqrt{3}} = 4\sqrt{3}$$3x^2 - y^2 = 16(3) = 48$$\frac{x^2}{16} - \frac{y^2}{48} = 1$This is a hyperbola with $a^2 = 16$ and $b^2 = 48$.The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{48}{16} = 1 + 3 = 4$.Therefore,$4e^2 = 4(4) = 16$.

The locus of the point of intersection of the lines $\sqrt{3}x - y - 4\sqrt{3}k = 0$ and $\sqrt{3}kx + ky - 4\sqrt{3} = 0$ for different real values of $k$ is a hyperbola $H$. If $e$ is the eccentricity of $H$,then $4e^2 =$

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