Q. For different values 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) an hyperbola with eccentricity $\sqrt{\frac{19}{16}}$
D) an ellipse with eccentricity $\frac{3}{4}$
Explanation
(a)
Equation of two straight lines are
$\sqrt{3}\ x - y = 4\sqrt{3}\alpha$
and $\sqrt{3}\ x + y = \frac{4\sqrt{3}}{\alpha}$
Solving above equations, we get
$3x^{2} - y^{2} = 48 \Rightarrow \ \frac{x^{2}}{16} - \frac{y^{2}}{48} = 1$
Which is a hyperbola
Whose eccentricity
$e = \sqrt{\frac{48 + 16}{16}} = \sqrt{4} = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.