If the slope of a tangent to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is `2sqrt(2)` then the eccentricity e of the hyperbola lies in the interval

Options

A. `1,sqrt(2)`
B. `1,2sqrt(2)`
C. `1,3`
D. `1,4`

Answer

To solve the problem, we need to find the eccentricity \( e \) of the hyperbola given the slope of the tangent line. The hyperbola is defined by the equation:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 1: Understand the relationship between the slope of the tangent and the hyperbola
The slope \( m \) of the tangent to the hyperbola at any point can be expressed as:

\[ m = \frac{b^2}{a^2} \cdot \frac{y}{x} \]

### Step 2: Substitute the given slope
We know that the slope of the tangent is given as \( m = 2\sqrt{2} \). Therefore, we can set up the equation:

\[ 2\sqrt{2} = \frac{b^2}{a^2} \cdot \frac{y}{x} \]

### Step 3: Express \( y \) in terms of \( x \)
Using the parametric equations of the hyperbola, we can express \( x \) and \( y \) in terms of a parameter \( \theta \):

\[ x = a \sec \theta, \quad y = b \tan \theta \]

Substituting these into the slope equation gives:

\[ 2\sqrt{2} = \frac{b^2}{a^2} \cdot \frac{b \tan \theta}{a \sec \theta} \]

### Step 4: Simplify the equation
This simplifies to:

\[ 2\sqrt{2} = \frac{b^3 \tan \theta}{a^3 \sec \theta} \]

Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \):

\[ 2\sqrt{2} = \frac{b^3 \sin \theta}{a^3 \cos^2 \theta} \]

### Step 5: Rearranging for \( \frac{b}{a} \)
Rearranging gives:

\[ \frac{b}{a} = 2\sqrt{2} \cdot \frac{\cos^2 \theta}{\sin \theta} \]

### Step 6: Find the eccentricity
The eccentricity \( e \) of the hyperbola is given by:

\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

Substituting \( \frac{b^2}{a^2} \):

\[ e = \sqrt{1 + \left(2\sqrt{2} \cdot \frac{\cos^2 \theta}{\sin \theta}\right)^2} \]

### Step 7: Simplify the expression for \( e \)
This becomes:

\[ e = \sqrt{1 + 8 \cdot \frac{\cos^4 \theta}{\sin^2 \theta}} \]

### Step 8: Determine the range of \( e \)
As \( \sin \theta \) varies from \( 0 \) to \( 1 \):

- When \( \sin \theta \to 0 \), \( e \to \infty \).
- When \( \sin \theta = 1 \), \( e = \sqrt{1 + 8} = 3 \).

Thus, the eccentricity \( e \) lies in the interval:

\[ (1, 3) \]

### Conclusion
The eccentricity \( e \) of the hyperbola lies in the interval \( (1, 3) \).