The length of the sub-tangent to the hyperbola $x^2 - 4y^2 = 4$ corresponding to the normal having slope unity is $\frac{1}{\sqrt{k}}$, then $k$ is equal to ______.
Step-by-Step Solution
Key Concept: For a hyperbola, if a normal has slope m, use the condition that the normal is perpendicular to the tangent (slope = -1/m). The sub-tangent is the horizontal distance from the foot of perpendicular from point of contact to the x-axis to the x-intercept of the tangent, calculated as |x₀|/|slope of tangent|.
The corresponding normals are $y = x \pm \frac{5\sqrt{3}}{3}$. For the normal $y = x - \sqrt{3}$, the corresponding tangent is $y = -x - \sqrt{3}$. The point of contact is $\left(-\frac{4}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$. The length of the sub-tangent is $NQ = NC - QC = \frac{4}{\sqrt{3}} - \sqrt{3} = \frac{1}{\sqrt{3}}$, giving $k = 3$.
Correct Answer: 3