Question

The locus of the point of intersection of the lines $\sqrt{3}x-y-4\sqrt{3}k=0\mathrm{and}\sqrt{3}kx+ky-4\sqrt{3}=0$ for different values of k is the hyperbola__________________________.

Answer 1

$$\begin{gathered} \mathrm{Since}\mathrm{given}\mathrm{lines}\mathrm{are} \\ \sqrt{3}x-y-4\sqrt{3k}=0\mathrm{i}.\mathrm{e}.k=\frac{\sqrt{3}x-y}{4\sqrt{3}} \\ \mathrm{and}\sqrt{3}kx+ky-4\sqrt{3}=0 \\ \mathrm{i}.\mathrm{e}.k=\frac{4\sqrt{3}}{\sqrt{3}x+y} \\ \mathrm{equating},\mathrm{the}\mathrm{value}\mathrm{of}k,\mathrm{we}\mathrm{get}, \\ \frac{\sqrt{3}x-y}{4\sqrt{3}}=\frac{4\sqrt{3}}{\sqrt{3}x+y} \\ \mathrm{i}.\mathrm{e}.\left(\sqrt{3}x-y\right)\left(\sqrt{3}x+y\right)=16\times 3 \\ \mathrm{i}.\mathrm{e}.3x^2-y^2=48 \\ \mathrm{i}.\mathrm{e}.\mathrm{locus}\mathrm{of}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{line}\mathrm{is}\mathrm{the}\mathrm{hyperbola}\mathrm{given}\mathrm{by}, \\ 3x^2-y^2=48 \end{gathered}$$