Question

The locus of the point of intersection of the lines \[ \sqrt3x-y-4\sqrt3k=0 \] and \[ \sqrt3kx+ky-4\sqrt3=0 \] for different real values of \(k\) is a hyperbola \(H\). If \(e\) is the eccentricity of \(H\), then \(4e^2=\)

• A. \(48\)
• B. \(39\)
• C. \(13\)
• D. \(16\)

Hint:

To find the locus involving a parameter, eliminate the parameter from the given equations and then compare the resulting equation with a standard conic form.

Answer 1

The Correct Option is D

Step 1: Eliminate k to find the locus.
From line 1: \(k = \dfrac{\sqrt{3}x-y}{4\sqrt{3}}\). From line 2: \(k = \dfrac{4\sqrt{3}}{\sqrt{3}x+y}\).

Step 2: Multiply the two expressions for k.
\[k^2 = \frac{(\sqrt{3}x-y) \cdot 4\sqrt{3}}{4\sqrt{3}(\sqrt{3}x+y)} = \frac{\sqrt{3}x-y}{\sqrt{3}x+y}\] Actually multiplying directly: \((\sqrt{3}x-y)(\sqrt{3}x+y) = 48\), so \(3x^2-y^2=48 \Rightarrow \dfrac{x^2}{16}-\dfrac{y^2}{48}=1\).

Step 3: Compute \(4e^2\).
\(e^2 = 1+\dfrac{48}{16} = 1+3 = 4\). So \(4e^2 = 16\).
\[ \boxed{16} \]