If the eccentricity of the hyperbola `x^(2) - y^(2) sec^(2)theta = 4` is `sqrt3` time the eccentricity of the ellipse `x^(2)sec^(2)theta+y^(2) = 16`, then the volue of `theta` equal

Options

A. `pi/6`
B. `(3pi)/4`
C. `pi/3`
D. `pi/2`

Answer

To solve the problem, we need to find the angle \( \theta \) given the relationship between the eccentricities of a hyperbola and an ellipse. We start by determining the eccentricities of both curves.

### Step 1: Identify the hyperbola and its eccentricity
The equation of the hyperbola is given as:
\[ x^2 - y^2 \sec^2 \theta = 4 \]
We can rewrite this in standard form:
\[ \frac{x^2}{4} - \frac{y^2}{4 \sec^2 \theta} = 1 \]
From this, we identify:
- \( a^2 = 4 \) → \( a = 2 \)
- \( b^2 = 4 \sec^2 \theta \) → \( b = 2 \sec \theta \)

The eccentricity \( e_h \) of the hyperbola is given by:
\[ e_h = \frac{c}{a} \quad \text{where } c = \sqrt{a^2 + b^2} \]
Calculating \( c \):
\[ c = \sqrt{4 + 4 \sec^2 \theta} = \sqrt{4(1 + \sec^2 \theta)} = 2\sqrt{1 + \sec^2 \theta} \]
Thus, the eccentricity becomes:
\[ e_h = \frac{2\sqrt{1 + \sec^2 \theta}}{2} = \sqrt{1 + \sec^2 \theta} \]

### Step 2: Identify the ellipse and its eccentricity
The equation of the ellipse is given as:
\[ x^2 \sec^2 \theta + y^2 = 16 \]
Rewriting this in standard form:
\[ \frac{x^2}{16 \sec^2 \theta} + \frac{y^2}{16} = 1 \]
From this, we identify:
- \( a^2 = 16 \) → \( a = 4 \)
- \( b^2 = 16 \sec^2 \theta \) → \( b = 4 \sec \theta \)

The eccentricity \( e_e \) of the ellipse is given by:
\[ e_e = \frac{c}{a} \quad \text{where } c = \sqrt{a^2 - b^2} \]
Calculating \( c \):
\[ c = \sqrt{16 - 16 \sec^2 \theta} = \sqrt{16(1 - \sec^2 \theta)} = 4\sqrt{1 - \sec^2 \theta} \]
Thus, the eccentricity becomes:
\[ e_e = \frac{4\sqrt{1 - \sec^2 \theta}}{4} = \sqrt{1 - \sec^2 \theta} \]

### Step 3: Set up the relationship between eccentricities
According to the problem, the eccentricity of the hyperbola is \( \sqrt{3} \) times the eccentricity of the ellipse:
\[ e_h = \sqrt{3} e_e \]
Substituting the expressions for \( e_h \) and \( e_e \):
\[ \sqrt{1 + \sec^2 \theta} = \sqrt{3} \sqrt{1 - \sec^2 \theta} \]

### Step 4: Square both sides
Squaring both sides gives:
\[ 1 + \sec^2 \theta = 3(1 - \sec^2 \theta) \]
Expanding and rearranging:
\[ 1 + \sec^2 \theta = 3 - 3\sec^2 \theta \]
\[ 4\sec^2 \theta = 2 \]
\[ \sec^2 \theta = \frac{1}{2} \]

### Step 5: Find \( \theta \)
Taking the reciprocal gives:
\[ \cos^2 \theta = 2 \quad \Rightarrow \quad \cos \theta = \pm \frac{1}{\sqrt{2}} \]
Thus, \( \theta \) can be:
\[ \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \]

### Step 6: Identify valid angles
From the options provided, we check which angle corresponds to \( \cos \theta = \pm \frac{1}{\sqrt{2}} \):
- \( \theta = \frac{3\pi}{4} \) is a valid solution.

Thus, the value of \( \theta \) is:
\[ \theta = \frac{3\pi}{4} \]