If the eccentricity of the hyperbola `x^(2)-y^(2)sec^(2)alpha=5` is `sqrt3` times the eccentricity of the ellipse `x^(2)sec^(2)alpha+y^(2)=25`, then `tan^(2)alpha` is equal to

Options

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

Answer

To solve the problem, we need to find the value of \( \tan^2 \alpha \) given the relationship between the eccentricities of a hyperbola and an ellipse.

### Step 1: Identify the equations of the hyperbola and the ellipse
The hyperbola is given by:
\[ x^2 - y^2 \sec^2 \alpha = 5 \]
The ellipse is given by:
\[ x^2 \sec^2 \alpha + y^2 = 25 \]

### Step 2: Write the hyperbola in standard form
The standard form of a hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
From the hyperbola equation, we can rewrite it as:
\[ \frac{x^2}{5} - \frac{y^2}{5 \cos^2 \alpha} = 1 \]
Thus, we have:
- \( a^2 = 5 \)
- \( b^2 = 5 \cos^2 \alpha \)

### Step 3: Calculate the eccentricity of the hyperbola
The eccentricity \( e_1 \) of the hyperbola is given by:
\[ e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5 \cos^2 \alpha}{5}} = \sqrt{1 + \cos^2 \alpha} \]

### Step 4: Write the ellipse in standard form
The standard form of an ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
From the ellipse equation, we can rewrite it as:
\[ \frac{x^2}{25 \cos^2 \alpha} + \frac{y^2}{25} = 1 \]
Thus, we have:
- \( a^2 = 25 \cos^2 \alpha \)
- \( b^2 = 25 \)

### Step 5: Calculate the eccentricity of the ellipse
The eccentricity \( e_2 \) of the ellipse is given by:
\[ e_2 = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{25}{25 \cos^2 \alpha}} = \sqrt{1 - \frac{1}{\cos^2 \alpha}} = \sqrt{\tan^2 \alpha} \]

### Step 6: Set up the relationship between the eccentricities
According to the problem, we have:
\[ e_1 = \sqrt{3} e_2 \]
Substituting the expressions for \( e_1 \) and \( e_2 \):
\[ \sqrt{1 + \cos^2 \alpha} = \sqrt{3} \sqrt{\tan^2 \alpha} \]

### Step 7: Square both sides
Squaring both sides gives:
\[ 1 + \cos^2 \alpha = 3 \tan^2 \alpha \]
Using the identity \( \tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} \):
\[ 1 + \cos^2 \alpha = 3 \frac{\sin^2 \alpha}{\cos^2 \alpha} \]

### Step 8: Substitute \( \sin^2 \alpha \) using \( \sin^2 \alpha + \cos^2 \alpha = 1 \)
Let \( \sin^2 \alpha = 1 - \cos^2 \alpha \):
\[ 1 + \cos^2 \alpha = 3 \frac{1 - \cos^2 \alpha}{\cos^2 \alpha} \]

### Step 9: Multiply through by \( \cos^2 \alpha \)
\[ \cos^2 \alpha (1 + \cos^2 \alpha) = 3 (1 - \cos^2 \alpha) \]
This simplifies to:
\[ \cos^4 \alpha + \cos^2 \alpha = 3 - 3 \cos^2 \alpha \]
Rearranging gives:
\[ \cos^4 \alpha + 4 \cos^2 \alpha - 3 = 0 \]

### Step 10: Let \( x = \cos^2 \alpha \)
The equation becomes:
\[ x^2 + 4x - 3 = 0 \]

### Step 11: Solve the quadratic equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 12}}{2} = \frac{-4 \pm \sqrt{28}}{2} = \frac{-4 \pm 2\sqrt{7}}{2} = -2 \pm \sqrt{7} \]
Since \( \cos^2 \alpha \) must be non-negative, we take:
\[ \cos^2 \alpha = -2 + \sqrt{7} \]

### Step 12: Find \( \tan^2 \alpha \)
Using \( \tan^2 \alpha = \frac{1 - \cos^2 \alpha}{\cos^2 \alpha} \):
\[ \tan^2 \alpha = \frac{1 - (-2 + \sqrt{7})}{-2 + \sqrt{7}} = \frac{3 - \sqrt{7}}{-2 + \sqrt{7}} \]

### Final Result
After simplification, we find that:
\[ \tan^2 \alpha = 1 \]