Question

The locus of the point of intersection of the lines (√3)kx + ky - 4√3 = 0 and √3 x - y - 4(√3)k = 0 is a conic, whose eccentricity is _________

Hint:

To find the locus of lines containing a parameter $k$, eliminate $k$ by substitution or multiplication.

Answer 1

Correct Answer: 2

Step 1: Eq 1: $k(\sqrt{3}x + y) = 4\sqrt{3}$.
Step 2: Eq 2: $k = \frac{\sqrt{3}x - y}{4\sqrt{3}}$.
Step 3: Multiply Eq 1 and Eq 2: $(\sqrt{3}x + y)(\frac{\sqrt{3}x - y}{4\sqrt{3}}) \cdot k = 4\sqrt{3} \cdot k$.
Step 4: $(\sqrt{3}x + y)(\sqrt{3}x - y) = (4\sqrt{3})^2 \Rightarrow 3x^2 - y^2 = 48$.
Step 5: $\frac{x^2}{16} - \frac{y^2}{48} = 1$. This is a hyperbola.
Step 6: $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{48}{16}} = \sqrt{1 + 3} = 2$.