Question

prove that the locus of the point of intersection of the lines x√3-y- 4k√3=0 and √3kx+ky-4√3=0 for different values of k is a hyperbola whose e is 2.

Answer 1

Dear student
$$\begin{gathered} \mathrm{The}\mathrm{equations}\sqrt{3}\mathrm{x}-\mathrm{y}-4\sqrt{3}\mathrm{k}=0\mathrm{and}\sqrt{3}\mathrm{kx}+\mathrm{ky}-4\sqrt{3}=0\mathrm{can}\mathrm{be}\mathrm{rewritten}\mathrm{as} \\ \sqrt{3}\mathrm{x}-\mathrm{y}=4\sqrt{3}\mathrm{k}\mathrm{and}\sqrt{3}\mathrm{kx}+\mathrm{ky}=4\sqrt{3}\mathrm{respectively}. \\ \mathrm{Multipying}\mathrm{the}\mathrm{equations}: \\ 3{\mathrm{kx}}^2-{\mathrm{ky}}^2=48\mathrm{k} \\ \Rightarrow \frac{3{\mathrm{kx}}^2}{48\mathrm{k}}-\frac{{\mathrm{ky}}^2}{48\mathrm{k}}=1 \\ \Rightarrow \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{48}=1 \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{standard}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{hyperbola},\mathrm{where}{\mathrm{a}}^2=16\mathrm{and}{\mathrm{b}}^2=48 \\ \mathrm{Eccentricity},\mathrm{e}=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2}} \\ \Rightarrow \mathrm{e}=\sqrt{\frac{16+48}{16}} \\ \Rightarrow \mathrm{e}=\frac{8}{4} \\ \Rightarrow \mathrm{e}=2 \end{gathered}$$
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