<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 to the hyperbola has slope 1, find the point of contact using the normal equation, then calculate the sub-tangent length using the formula: sub-tangent = y/(dy/dx) at that point.
<p><strong>Step 1:</strong> Rewrite hyperbola in standard form: $\frac{x^2}{4} - \frac{y^2}{1} = 1$, so $a^2 = 4, b^2 = 1$.</p><p><strong>Step 2:</strong> Differentiate implicitly: $2x - 8y\frac{dy}{dx} = 0$, giving $\frac{dy}{dx} = \frac{x}{4y}$</p><p><strong>Step 3:</strong> Slope of normal = $-\frac{1}{dy/dx} = -\frac{4y}{x} = 1$ (given). Therefore: $-4y = x$, or $x = -4y$</p><p><strong>Step 4:</strong> Substitute into hyperbola equation: $(-4y)^2 - 4y^2 = 4 \Rightarrow 16y^2 - 4y^2 = 4 \Rightarrow 12y^2 = 4 \Rightarrow y^2 = \frac{1}{3}$</p><p><strong>Step 5:</strong> So $y = \pm\frac{1}{\sqrt{3}}$ and $x = \mp\frac{4}{\sqrt{3}}$</p><p><strong>Step 6:</strong> At this point, $\frac{dy}{dx} = \frac{x}{4y} = \frac{-4y}{4y} = -1$ (verify slope of tangent)</p><p><strong>Step 7:</strong> Sub-tangent length = $\left|\frac{y}{dy/dx}\right| = \left|\frac{1/\sqrt{3}}{-1}\right| = \frac{1}{\sqrt{3}}$</p><p><strong>Step 8:</strong> Given sub-tangent = $\frac{1}{\sqrt{k}}$, we have $\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{k}}$</p><p>∴ Answer: $k = 3$ (Option C)</p>
Correct Answer: C