Question

The locus of the point of intersection on the line \( \sqrt{3}x - y - 4\sqrt{3}k = 0 \) and \( \sqrt{3}kx + ky - 4\sqrt{3} = 0 \) 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 of the point of intersection of two lines involving a parameter (k), eliminate k from the two equations.
If the equations are \(L_1 + kL_2 = 0\) and \(L_3 + kL_4 = 0\), then \(L_1 L_4 - L_2 L_3 = 0\) is the locus (family of conics passing through intersection of \(L_1,L_3\); \(L_1,L_4\); etc.).
Here, it was simpler: express k from one and substitute, or make terms \(X=f(k)\) and \(Y=g(k)\) and eliminate k to find relation between X and Y.
Standard hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Eccentricity \(e = \sqrt{1+b^2/a^2}\).

Answer 1

The Correct Option is D

The given lines are: L1: \( \sqrt{3}x - y = 4\sqrt{3}k \) L2: \( \sqrt{3}kx + ky = 4\sqrt{3} \implies k(\sqrt{3}x + y) = 4\sqrt{3} \implies \sqrt{3}x + y = \frac{4\sqrt{3}}{k} \) (assuming \(k \neq 0\)). Let \(X = \sqrt{3}x - y\) and \(Y = \sqrt{3}x + y\). Then from L1, \(X = 4\sqrt{3}k\). From L2, \(Y = \frac{4\sqrt{3}}{k}\). Multiplying these two equations: \(XY = (4\sqrt{3}k) \left(\frac{4\sqrt{3}}{k}\right) = (4\sqrt{3})^2 = 16 \times 3 = 48\). So, \((\sqrt{3}x - y)(\sqrt{3}x + y) = 48\). This is of the form \((A-B)(A+B) = A^2-B^2\). \((\sqrt{3}x)^2 - y^2 = 48\) \(3x^2 - y^2 = 48\). Divide by 48 to get standard hyperbola form: \( \frac{3x^2}{48} - \frac{y^2}{48} = 1 \) \( \frac{x^2}{16} - \frac{y^2}{48} = 1 \). This is a hyperbola of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, \(a^2 = 16\) and \(b^2 = 48\). The eccentricity \(e\) of a hyperbola is given by \(e = \sqrt{1 + \frac{b^2}{a^2}}\). \(e = \sqrt{1 + \frac{48}{16}} = \sqrt{1 + 3} = \sqrt{4} = 2\). We need to find \(4e^2\). \(e^2 = 4\). \(4e^2 = 4 \times 4 = 16\). This matches option (d). \[ \boxed{16} \]