Question

The locus of the point of intersection of the lines $\sqrt{3}x-y-4\sqrt{3}\mathrm{\lambda }=0\mathrm{and}\sqrt{3}\mathrm{\lambda }x+\mathrm{\lambda }y-4\sqrt{3}=0$ is a hyperbola of eccentricity
(a) 1
(b) 2
(c) 3
(d) 4

Answer 1

(b) 2

The equations of lines $\sqrt{3}x-y-4\sqrt{3}\lambda =0\mathrm{and}\sqrt{3}\lambda x+\lambda y-4\sqrt{3}=0$ can be rewritten as $\sqrt{3}x-y=4\sqrt{3}\lambda \mathrm{and}\sqrt{3}\lambda x+\lambda y=4\sqrt{3}$, respectively.

Multiplying the equations:

$$\begin{gathered} 3\lambda x^2-\lambda y^2=48\lambda \\ \Rightarrow \frac{3\lambda x^2}{48\lambda }-\frac{\lambda y^2}{48\lambda }=1 \\ \Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1 \end{gathered}$$
This is the standard equation of a hyperbola, where $a^2=16\mathrm{and}{b}^2=48$.

$$\begin{gathered} \mathrm{Eccentricity},e=\sqrt{\frac{a^2+b^2}{a^2}} \\ \Rightarrow e=\sqrt{\frac{16+48}{16}} \\ \Rightarrow e=\frac{8}{4} \\ \Rightarrow e=2 \end{gathered}$$