Question

For different valus of $\alpha$, the locus of the point of intersection of the two straight lines $\sqrt{3}x-y-4\sqrt{3}\alpha =0$ and $\sqrt{3}\alpha x+\alpha y-4\sqrt{3}=0$ is

• A. A hyperbola with eccentricity $2$
• B. An ellipse with eccentricity $\sqrt{\frac{2}{3}}$
• C. A hyperbola with eccentricity $\sqrt{\frac{19}{16}}$
• D. An ellipse with eccentricity $\frac{3}{4}$

Answer 1

The correct option is A A hyperbola with eccentricity $2$

Given, $\sqrt{3}x-y=-4\sqrt{3}\alpha =0$
$\Rightarrow \sqrt{3}x-y=4\sqrt{3}\alpha \dots \left(i\right)$
and $x\sqrt{3}\alpha +\alpha y-4\sqrt{3}=0$
$\Rightarrow \sqrt{3}x+y=\frac{4\sqrt{3}\alpha }{}\dots \left(ii\right)$
On multiplying Eq. (i) by Eq. (ii)
$\Rightarrow 3x^2-y^2=48$
$\Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1$
Which is hyperbola.
Hence, eccentricity $e=\sqrt{\frac{48+16}{16}}=2$