Question

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

• A. 2.00
• B. 2
• C. 2.0
• D. 02

Answer 1

Answer: (A), (B), (C), (D)

$\sqrt{3}kx+ky=4\sqrt{3}......\left(1\right)$
$\sqrt{3}kx-ky=4\sqrt{3}{k}^2.....\left(2\right)$
Adding equation $\left(1\right)$ and $\left(2\right)$
$2\sqrt{3}kx=4\sqrt{3}(k^2+1)$
$x=2(k+\frac{1}{k})....\left(3\right)$
Substracting equation $\left(1\right)$ and $\left(2\right)$
$y=2\sqrt{3}(\frac{1}{k}-k)....\left(4\right)$
By $\left(3\right)$ and $\left(4\right)$
$\therefore \frac{x^2}{4}-\frac{y^2}{12}=4$
$\frac{x^2}{16}-\frac{y^2}{48}=1$ Hyperbola
$\therefore e^2=1+\frac{48}{16}$
$e=2$