Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

We have,

$x\sqrt{3}-y-4\sqrt{3}k=0.......\left(1\right)$

$kx\sqrt{3}+ky-4\sqrt{3}=0.......\left(2\right)$

By equation (1) and we get,

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

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

$k=\frac{x\sqrt{3}-y}{4\sqrt{3}}........\left(3\right)$

Now, by equation (2) and we get,

$kx\sqrt{3}+ky-4\sqrt{3}=0$

$kx\sqrt{3}+ky=4\sqrt{3}$

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

$k=\frac{4\sqrt{3}}{(x\sqrt{3}+y)}.......\left(4\right)$

By equation (3) and (4) to and we get,

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

${\left(x\sqrt{3}\right)}^2-y^2={\left(4\sqrt{3}\right)}^2$

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

$\frac{3x^2-y^2}{48}=1$

$\frac{3x^2}{48}-\frac{y^2}{48}=1$

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

Hence, this equation is hyperbola.

Then its eccentricity.

Comparing that equation (5) to

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Now,

$a^2=16,b^2=48$

Then,

$Eccentricity\left(e\right)=\sqrt{1+\frac{b^2}{a^2}}$

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

$e=\sqrt{\frac{64}{16}}$

$e=\sqrt{4}$

$e=2$

Hence, this is the answer.