For different values of alpha, the locus of t | vedclass

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(A) Given equations are:$L_1: \sqrt{3} x - y = 4 \sqrt{3} \alpha$  $(1)$$L_2: \alpha(\sqrt{3} x + y) = 4 \sqrt{3} \Rightarrow \sqrt{3} x + y = \frac{4

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a hyperbola with eccentricity $2$

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(A) Given equations are:$L_1: \sqrt{3} x - y = 4 \sqrt{3} \alpha$  $(1)$$L_2: \alpha(\sqrt{3} x + y) = 4 \sqrt{3} \Rightarrow \sqrt{3} x + y = \frac{4 \sqrt{3}}{\alpha}$  $(2)$Multiplying equation $(1)$ and $(2)$, we get:$(\sqrt{3} x - y)(\sqrt{3} x + y) = (4 \sqrt{3} \alpha) 	imes \left(\frac{4 \sqrt{3}}{\alpha}ight)$$3x^2 - y^2 = 16 	imes 3 = 48$Dividing by $48$, we get:$\frac{x^2}{16} - \frac{y^2}{48} = 1$This is the equation of a hyperbola where $a^2 = 16$ and $b^2 = 48$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{48}{16}} = \sqrt{1 + 3} = \sqrt{4} = 2$.Thus, it is a hyperbola with eccentricity $2$.

For different values of $\alpha$, the locus of the point of intersection of the two straight lines $\sqrt{3} x - y - 4 \sqrt{3} \alpha = 0$ and $\sqrt{3} \alpha x + \alpha y - 4 \sqrt{3} = 0$ is

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