A hyperbola having the transverse axis of length `(1)/(2)` unit is confocal with the ellipse `3x^(2)+4y^(2) = 12`, then

Options

A. Equation of the hyperbola is `(x^(2))/(15)-(y^(2))/(1) = (1)/(16)`
B. Eccentricity of the hyperbola is 4
C. Distance between the directries of the hyperbola is `(1)/(8)` units
D. Length of latus rectum of the hyperbola is `(15)/(2)` units

Explanation

To solve the problem step by step, we will analyze the given information about the hyperbola and the confocal ellipse.

### Step 1: Write the equation of the ellipse in standard form.
The given equation of the ellipse is:
\[ 3x^2 + 4y^2 = 12 \]

To convert this into standard form, we divide the entire equation by 12:
\[ \frac{3x^2}{12} + \frac{4y^2}{12} = 1 \]
This simplifies to:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]

### Step 2: Identify the values of \( a^2 \) and \( b^2 \).
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can see that:
- \( a^2 = 4 \) (thus \( a = 2 \))
- \( b^2 = 3 \) (thus \( b = \sqrt{3} \))

### Step 3: Calculate the eccentricity of the ellipse.
The formula for the eccentricity \( e \) of an ellipse is given by:
\[ e^2 = 1 - \frac{b^2}{a^2} \]
Substituting the values we found:
\[ e^2 = 1 - \frac{3}{4} = \frac{1}{4} \]
Thus, the eccentricity \( e \) is:
\[ e = \frac{1}{2} \]

### Step 4: Determine the foci of the ellipse.
The foci of the ellipse are given by \( (\pm ae, 0) \):
\[ \text{Foci} = (\pm 2 \cdot \frac{1}{2}, 0) = (\pm 1, 0) \]

### Step 5: Understand the confocal hyperbola.
Since the hyperbola is confocal with the ellipse, it will have the same foci \( (\pm 1, 0) \).

### Step 6: Use the transverse axis length to find \( a \) for the hyperbola.
The transverse axis of the hyperbola is given as \( \frac{1}{2} \) units. Therefore, we have:
\[ 2a = \frac{1}{2} \]
This implies:
\[ a = \frac{1}{4} \]

### Step 7: Calculate the eccentricity of the hyperbola.
For hyperbolas, the relationship between \( a \), \( b \), and eccentricity \( e_1 \) is given by:
\[ e_1^2 = 1 + \frac{b^2}{a^2} \]
We can rearrange this to find \( b^2 \):
\[ b^2 = a^2(e_1^2 - 1) \]

### Step 8: Find \( e_1 \) using the foci.
Since the foci of the hyperbola are also \( (\pm 1, 0) \), we have:
\[ ae_1 = 1 \]
Substituting \( a = \frac{1}{4} \):
\[ \frac{1}{4} e_1 = 1 \]
Thus:
\[ e_1 = 4 \]

### Step 9: Calculate \( b^2 \).
Now substituting \( a = \frac{1}{4} \) and \( e_1 = 4 \):
\[ b^2 = \left(\frac{1}{4}\right)^2 (16 - 1) = \frac{1}{16} \cdot 15 = \frac{15}{16} \]

### Step 10: Write the equation of the hyperbola.
The standard form of the hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Substituting \( a^2 = \frac{1}{16} \) and \( b^2 = \frac{15}{16} \):
\[ \frac{x^2}{\frac{1}{16}} - \frac{y^2}{\frac{15}{16}} = 1 \]
This simplifies to:
\[ 16x^2 - 15y^2 = 1 \]

### Final Step: Summary of results.
Thus, the equation of the hyperbola is:
\[ 16x^2 - 15y^2 = 1 \]