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(D) The given equations are:$\sqrt{3}x - y = 4\sqrt{3}k$  $(i)$$k(\sqrt{3}x + y) = 4\sqrt{3}$  $(ii)$To find the locus of the point of intersection,we

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Hyperbola

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(D) The given equations are:$\sqrt{3}x - y = 4\sqrt{3}k$  $(i)$$k(\sqrt{3}x + y) = 4\sqrt{3}$  $(ii)$To find the locus of the point of intersection,we eliminate $k$ by multiplying the two equations:$(\sqrt{3}x - y) 	imes k(\sqrt{3}x + y) = (4\sqrt{3}k) 	imes (4\sqrt{3})$$(\sqrt{3}x - y)(\sqrt{3}x + y) = 16 	imes 3$$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.

Find the locus of the point of intersection of the lines $\sqrt{3}x - y - 4\sqrt{3}k = 0$ and $\sqrt{3}kx + yk - 4\sqrt{3} = 0$ for different values of $k$.

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