Let `0 lt theta lt (pi)/2`. If the eccentricity of the hyperbola `(x^(2))/(cos^(2)theta)-(y^(2))/(sin^(2)theta)=1` is greater ten 2, then the length of its latus rectum lies in the interval,

Options

A. `(3,oo)`
B. `(3/2,2]`
C. `(1,3/2]`
D. `(2,3]`

Explanation

To solve the problem step by step, we will analyze the given hyperbola and derive the necessary conditions for its eccentricity and the length of the latus rectum.

### Step 1: Understand the hyperbola equation
The hyperbola is given by the equation:

\[ \frac{x^2}{\cos^2 \theta} - \frac{y^2}{\sin^2 \theta} = 1 \]

### Step 2: Identify the eccentricity
For a hyperbola of the form:

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

the eccentricity \( e \) is given by:

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

In our case, \( a^2 = \cos^2 \theta \) and \( b^2 = \sin^2 \theta \). Thus, we can express the eccentricity as:

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

### Step 3: Set up the inequality for eccentricity
We are given that the eccentricity is greater than 2:

\[ e > 2 \]

Substituting the expression for \( e \):

\[ \sqrt{1 + \tan^2 \theta} > 2 \]

### Step 4: Square both sides
Squaring both sides (since both sides are positive):

\[ 1 + \tan^2 \theta > 4 \]

### Step 5: Rearrange the inequality
Rearranging gives:

\[ \tan^2 \theta > 3 \]

Taking the square root:

\[ \tan \theta > \sqrt{3} \]

### Step 6: Determine the range for theta
The inequality \( \tan \theta > \sqrt{3} \) implies:

\[ \theta > \frac{\pi}{3} \]

Since \( 0 < \theta < \frac{\pi}{2} \), we have:

\[ \frac{\pi}{3} < \theta < \frac{\pi}{2} \]

### Step 7: Find the length of the latus rectum
The length of the latus rectum \( L \) for the hyperbola is given by:

\[ L = \frac{2b^2}{a} \]

Substituting \( b^2 = \sin^2 \theta \) and \( a = \cos \theta \):

\[ L = \frac{2 \sin^2 \theta}{\cos \theta} \]

### Step 8: Analyze the limits of L
Now we need to analyze the limits of \( L \) as \( \theta \) varies from \( \frac{\pi}{3} \) to \( \frac{\pi}{2} \).

1. When \( \theta = \frac{\pi}{3} \):

\[ L = \frac{2 \sin^2 \frac{\pi}{3}}{\cos \frac{\pi}{3}} = \frac{2 \left(\frac{\sqrt{3}}{2}\right)^2}{\frac{1}{2}} = \frac{2 \cdot \frac{3}{4}}{\frac{1}{2}} = \frac{3}{2} \cdot 2 = 3 \]

2. When \( \theta \to \frac{\pi}{2} \):

As \( \theta \) approaches \( \frac{\pi}{2} \), \( \cos \theta \) approaches 0, thus \( L \) approaches infinity.

### Final Result
Thus, the length of the latus rectum \( L \) lies in the interval:

\[ (3, \infty) \]