Question

The locus of the point of intersection of the lines $\sqrt{3}x-y-4\sqrt{3}\lambda =0$ and $\sqrt{3}\lambda x+\lambda y-4\sqrt{3}=0$ is a hyperbola of eccentricity

Answer 1

Let point of intersection of lines is $(h,k)$

So, the point $(h,k)$ lies on both lines.

$\therefore \sqrt{3}x-y-4\sqrt{3}\lambda =0$

$\Rightarrow \sqrt{3}h-k-4\sqrt{3}\lambda =0$

$\Rightarrow \lambda =\frac{\sqrt{3}h-k}{4\sqrt{3}}\dots \left(1\right)$

And $\sqrt{3}\lambda x+\lambda y-4\sqrt{3}=0$

$\Rightarrow \sqrt{3}\lambda h+\lambda k-4\sqrt{3}=0$

$\Rightarrow \sqrt{3}h\times \frac{\sqrt{3}h-k}{4\sqrt{3}}+k\times \frac{\sqrt{3}h-k}{4\sqrt{3}}-4\sqrt{3}=0$
(from $\left(1\right)$)

$\Rightarrow 3h^2-\sqrt{3}hk+\sqrt{3}hk-k^2-48=0$

$\Rightarrow 3h^2-k^2=48$

For the locus replace $(h,k)$ by $(x,y)$

$\Rightarrow 3x^2-y^2=48\dots \left(2\right)$

Hence, the locus is an equation of hyperbola. Now, finding its eccentricity,

$\frac{x^2}{16}-\frac{y^2}{48}=1$ (from (2))

$\therefore a^2=16$ and $b^2=48$

$\therefore e=\sqrt{1+\frac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1+\frac{48}{16}}$

$\Rightarrow e=\sqrt{1+3}$

$\Rightarrow e=2$

Hence, option $\left(B\right)$ is correct.