The length of the sub-tangent to the hyperbola `x^(2)-4y^(2)=4` corresponding to the normal having slope unity is `(1)/(sqrtk),` then the value of k is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem step by step, we will follow the given information about the hyperbola and the properties of tangents and normals.

### Step 1: Identify the Hyperbola
The equation of the hyperbola is given as:
\[ x^2 - 4y^2 = 4 \]
We can rewrite this in standard form:
\[ \frac{x^2}{4} - \frac{y^2}{1} = 1 \]

### Step 2: Determine the Slope of the Tangent
We are given that the slope of the normal is unity (1). The slope of the tangent \(m_t\) is related to the slope of the normal \(m_n\) by the relation:
\[ m_t \cdot m_n = -1 \]
Thus, if \(m_n = 1\), then:
\[ m_t = -1 \]

### Step 3: Find the Equation of the Tangent
The equation of the tangent to the hyperbola at the point \((x_0, y_0)\) is given by:
\[ \frac{xx_0}{4} - \frac{yy_0}{1} = 1 \]
Since we know the slope of the tangent is -1, we can express the tangent line in slope-intercept form:
\[ y - y_0 = -1(x - x_0) \implies y = -x + (y_0 + x_0) \]

### Step 4: Find the Point on the Hyperbola
To find the point on the hyperbola where the slope of the tangent is -1, we can substitute \(y\) from the tangent equation into the hyperbola equation. However, we can also use the implicit differentiation of the hyperbola to find the slope.

Differentiating the hyperbola implicitly:
\[ 2x - 8y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{4y} \]
Setting this equal to -1 (the slope of the tangent):
\[ \frac{x}{4y} = -1 \implies x = -4y \]

### Step 5: Substitute into the Hyperbola Equation
Substituting \(x = -4y\) into the hyperbola equation:
\[ (-4y)^2 - 4y^2 = 4 \implies 16y^2 - 4y^2 = 4 \implies 12y^2 = 4 \implies y^2 = \frac{1}{3} \]
Thus:
\[ y = \pm \frac{1}{\sqrt{3}} \]

### Step 6: Find Corresponding \(x\) Values
Using \(y = \frac{1}{\sqrt{3}}\):
\[ x = -4\left(\frac{1}{\sqrt{3}}\right) = -\frac{4}{\sqrt{3}} \]
And for \(y = -\frac{1}{\sqrt{3}}\):
\[ x = -4\left(-\frac{1}{\sqrt{3}}\right) = \frac{4}{\sqrt{3}} \]
So the points on the hyperbola are:
\[ \left(-\frac{4}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \text{ and } \left(\frac{4}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) \]

### Step 7: Length of the Normal
The length of the normal at any point \((x_0, y_0)\) on the hyperbola can be calculated using the formula:
\[ \text{Length of Normal} = y \cdot \frac{dx}{dy} \]
From our earlier calculations, we have:
\[ \frac{dx}{dy} = 1 \text{ (since slope of normal is 1)} \]
Thus:
\[ \text{Length of Normal} = y \cdot 1 = y \]
For \(y = \frac{1}{\sqrt{3}}\):
\[ \text{Length of Normal} = \frac{1}{\sqrt{3}} \]

### Step 8: Relate to Given Length
We are given that the length of the sub-tangent is \(\frac{1}{\sqrt{k}}\). Therefore:
\[ \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{k}} \implies \sqrt{k} = \sqrt{3} \implies k = 3 \]

### Final Answer
The value of \(k\) is:
\[ \boxed{3} \]