<p>The length of sub-tangent to the hyperbola \(x^2 - 4y^2 = 4\) corresponding to the normal having slope unity is \(\dfrac{1}{\sqrt{k}}\), then the value of \(k\) is:</p>
Step-by-Step Solution
Key Concept: If a normal has slope 1, the tangent has slope -1. Use the tangent slope formula for hyperbola to find the point of contact, then calculate the sub-tangent length using the formula: sub-tangent = y/(dy/dx) evaluated at that point.
<p><strong>Step 1:</strong> Rewrite the hyperbola as $\frac{x^2}{4} - y^2 = 1$, so $a^2 = 4, b^2 = 1$.</p><p><strong>Step 2:</strong> If the normal has slope 1, the tangent has slope -1 (since slopes of perpendicular lines multiply to -1).</p><p><strong>Step 3:</strong> For the hyperbola $\frac{x^2}{4} - y^2 = 1$, differentiate implicitly: $\frac{x}{2} - 2y\frac{dy}{dx} = 0$, giving $\frac{dy}{dx} = \frac{x}{4y}$.</p><p><strong>Step 4:</strong> At the point of tangency, $\frac{dy}{dx} = -1$, so $\frac{x}{4y} = -1 \Rightarrow x = -4y$.</p><p><strong>Step 5:</strong> Substitute $x = -4y$ into the hyperbola equation: $\frac{(-4y)^2}{4} - y^2 = 1 \Rightarrow 4y^2 - y^2 = 1 \Rightarrow 3y^2 = 1 \Rightarrow y = \pm\frac{1}{\sqrt{3}}$.</p><p><strong>Step 6:</strong> The sub-tangent is defined as $\left|\frac{y}{dy/dx}\right| = \left|\frac{y}{-1}\right| = |y| = \frac{1}{\sqrt{3}}$.</p><p><strong>Step 7:</strong> Given that sub-tangent $= \frac{1}{\sqrt{k}}$, we have $\frac{1}{\sqrt{k}} = \frac{1}{\sqrt{3}}$.</p><p>∴ <strong>k = 3</strong></p>
Correct Answer: C