Question

The locus of the point of intersection of the lines $\left(\sqrt{3}\right)kx+ky-4\sqrt{3}=0$ and $\sqrt{3}x–y–4\sqrt{3}k=0$is a conic, whose eccentricity is __________.

Answer 1

Compute the eccentricity.

Given:

$\sqrt{3}kx+ky-4\sqrt{3}=0$…………………$\left(1\right)$

$\sqrt{3}x–y–4\sqrt{3}k=0$

Multiply with $k$ ,

$\sqrt{3}kx–ky–4\sqrt{3}{k}^2=0$…….………….$\left(2\right)$

Add equation $\left(1\right)$ and $\left(2\right)$,

$2\sqrt{3}kx–4\sqrt{3}\left[1+k^2\right]=0$

$\Rightarrow$ $2\sqrt{3}kx=4\sqrt{3}\left[1+k^2\right]$

$\Rightarrow$ $x=2\left[k+\frac{1}{k}\right]$……………$\left(3\right)$

Subtract equation $\left(1\right)$ and $\left(2\right)$

$2ky–4\sqrt{3}\left(1+k^2\right)=0$

$\Rightarrow$ $2ky=4\sqrt{3}\left(1+k^2\right)$

$\Rightarrow$ $y=2\sqrt{3}\left(\frac{1}{k}-k\right)$…………….$\left(4\right)$

From equation $\left(3\right)$ and $\left(4\right)$,

$\frac{x^2}{4}-\frac{y^2}{12}=k^2+{\left(\frac{1}{k}\right)}^2+2-{\left(\frac{1}{k}\right)}^2-k^2+2$

$\Rightarrow$$\frac{x^2}{4}-\frac{y^2}{12}=4$

$\Rightarrow$$\frac{x^2}{16}-\frac{y^2}{48}=1$………………$\left(5\right)$

The equation $\left(5\right)$ is the equation of the hyperbola.

Therefore, eccentricity is,

$e^2=1+\frac{48}{16}$

$\Rightarrow$$e^2=4$

$\Rightarrow$ $e=2$

Hence, eccentricity is $2$.