It the eccentricity of the hyerbola
`x^(2)-y^(2)cos ce^(2) alpha = 25` is `sqrt5` time the eccentricity of the ellipse `x^(2)cos ce^(2) alpha + y^(2) = 5`, then `alpha` is equal to :

Options

A. `tan^(-1) sqrt2`
B. `sin^(-1)sqrt(3/4)`
C. `tan^(-1)sqrt(2/5)`
D. `sin^(-1)sqrt(2/5)`

Answer

To solve the problem, we need to find the value of angle \( \alpha \) given the relationship between the eccentricities of a hyperbola and an ellipse. Let's break down the solution step by step.

### Step 1: Identify the equations of the conics

The hyperbola is given by:
\[ x^2 - y^2 \cos^2 \alpha = 25 \]
We can rewrite this in standard form:
\[ \frac{x^2}{25} - \frac{y^2}{25 \sin^2 \alpha} = 1 \]

The ellipse is given by:
\[ x^2 \cos^2 \alpha + y^2 = 5 \]
We can rewrite this in standard form:
\[ \frac{x^2}{5 \sin^2 \alpha} + \frac{y^2}{5} = 1 \]

### Step 2: Determine \( a^2 \) and \( b^2 \) for the hyperbola

From the hyperbola equation, we have:
- \( a^2 = 25 \) so \( a = 5 \)
- \( b^2 = 25 \sin^2 \alpha \) so \( b = 5 \sin \alpha \)

### Step 3: Calculate the eccentricity of the hyperbola

The eccentricity \( e_h \) of the hyperbola is given by:
\[ e_h = \frac{c}{a} \]
where \( c = \sqrt{a^2 + b^2} \).

Calculating \( c \):
\[ c = \sqrt{25 + 25 \sin^2 \alpha} = 5 \sqrt{1 + \sin^2 \alpha} \]
Thus, the eccentricity becomes:
\[ e_h = \frac{5 \sqrt{1 + \sin^2 \alpha}}{5} = \sqrt{1 + \sin^2 \alpha} \]

### Step 4: Determine \( a^2 \) and \( b^2 \) for the ellipse

From the ellipse equation, we have:
- \( a^2 = 5 \sin^2 \alpha \) so \( a = \sqrt{5} \)
- \( b^2 = 5 \) so \( b = \sqrt{5} \)

### Step 5: Calculate the eccentricity of the ellipse

The eccentricity \( e_e \) of the ellipse is given by:
\[ e_e = \frac{c}{a} \]
where \( c = \sqrt{a^2 - b^2} \).

Calculating \( c \):
\[ c = \sqrt{5 - 5 \sin^2 \alpha} = \sqrt{5(1 - \sin^2 \alpha)} = \sqrt{5 \cos^2 \alpha} = \sqrt{5} \cos \alpha \]
Thus, the eccentricity becomes:
\[ e_e = \frac{\sqrt{5} \cos \alpha}{\sqrt{5 \sin^2 \alpha}} = \frac{\cos \alpha}{\sin \alpha} = \cot \alpha \]

### Step 6: Set up the relationship between the eccentricities

According to the problem, the eccentricity of the hyperbola is \( \sqrt{5} \) times the eccentricity of the ellipse:
\[ \sqrt{1 + \sin^2 \alpha = \sqrt{5} \cdot \cot \alpha} \]

### Step 7: Square both sides to eliminate the square root

Squaring both sides gives:
\[ 1 + \sin^2 \alpha = 5 \cdot \cot^2 \alpha \]
Using \( \cot^2 \alpha = \frac{\cos^2 \alpha}{\sin^2 \alpha} \):
\[ 1 + \sin^2 \alpha = 5 \cdot \frac{\cos^2 \alpha}{\sin^2 \alpha} \]

### Step 8: Substitute \( \cos^2 \alpha = 1 - \sin^2 \alpha \)

Substituting \( \cos^2 \alpha \):
\[ 1 + \sin^2 \alpha = 5 \cdot \frac{1 - \sin^2 \alpha}{\sin^2 \alpha} \]
Multiplying through by \( \sin^2 \alpha \):
\[ \sin^2 \alpha + \sin^4 \alpha = 5(1 - \sin^2 \alpha) \]
Rearranging gives:
\[ \sin^4 \alpha + 6 \sin^2 \alpha - 5 = 0 \]

### Step 9: Let \( x = \sin^2 \alpha \)

This leads to a quadratic equation:
\[ x^2 + 6x - 5 = 0 \]
Using the quadratic formula:
\[ x = \frac{-6 \pm \sqrt{36 + 20}}{2} = \frac{-6 \pm \sqrt{56}}{2} = \frac{-6 \pm 2\sqrt{14}}{2} = -3 \pm \sqrt{14} \]

### Step 10: Find \( \sin^2 \alpha \)

Since \( \sin^2 \alpha \) must be non-negative:
\[ \sin^2 \alpha = -3 + \sqrt{14} \]

### Step 11: Find \( \alpha \)

Finally, we can find \( \alpha \) using:
\[ \alpha = \sin^{-1}(\sqrt{-3 + \sqrt{14}}) \]