Question

The locus of the point of intersection of the lines $\sqrt{3x}-y-4\sqrt{3}t=0\&\sqrt{3}tx+ty-4\sqrt{3}=0$ (where $t$ is a parameter) is a hyperbola. whose eccentricity is

• A. $\sqrt{3}$
• B. $2$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{4}{3}$

Answer 1

Answer: (B)

$2$

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

$\sqrt{3}x-y=4\sqrt{3}t$

$\Rightarrow t=\frac{\sqrt{3}x-y}{4\sqrt{3}}$ ............. (1)

$\sqrt{3}tx+ty-4\sqrt{3}=0$

$t(\sqrt{3}x+y)=4\sqrt{3}$

$\Rightarrow t=\frac{4\sqrt{3}}{\sqrt{3}x+y}$ ............. (2)

From (1) and (2), The locus of the point of intersection of the lines is,

$\frac{\sqrt{3}x-y}{4\sqrt{3}}=\frac{4\sqrt{3}}{\sqrt{3}x+y}$

$3x^2-y^2=48$

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

The above equation is a hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Eccentricity, $e=\frac{\sqrt{a^2+b^2}}{a}$

$e=\frac{\sqrt{16+48}}{4}=\frac{\sqrt{64}}{4}=2$