A hyperbola having the transverse axis of length `sqrt2` units has the same focii as that of ellipse `3x^(2)+4y^(2)=12`, then its equation is

Options

A. `2x^(2)-2y^(2)=1`
B. `2x^(2)-2y^(2)=3`
C. `x^(2)-y^(2)=-2`
D. `x^(2)-y^(2)=2`

Answer

To solve the problem, we need to find the equation of a hyperbola that has the same foci as a given ellipse. The steps to derive the equation are as follows:

### Step 1: Identify the Ellipse Parameters
The equation of the ellipse is given as:
\[ 3x^2 + 4y^2 = 12 \]

We can rewrite this in standard form:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]

From this, we can identify:
- \( a^2 = 4 \) (thus, \( a = 2 \))
- \( b^2 = 3 \) (thus, \( b = \sqrt{3} \))

### Step 2: Calculate the Eccentricity of the Ellipse
The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Substituting the values:
\[ e = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 3: Find the Foci of the Ellipse
The foci of the ellipse are located at:
\[ (\pm ae, 0) \]
Substituting the values of \( a \) and \( e \):
\[ (\pm 2 \cdot \frac{1}{2}, 0) = (\pm 1, 0) \]

### Step 4: Determine the Transverse Axis of the Hyperbola
The problem states that the transverse axis of the hyperbola is \( \sqrt{2} \). The length of the transverse axis is given by \( 2a \):
\[ 2a = \sqrt{2} \]
Thus, we find:
\[ a = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]

### Step 5: Calculate the Eccentricity of the Hyperbola
For the hyperbola, the foci are also given by:
\[ (\pm ae, 0) \]
Since the foci of the hyperbola are the same as those of the ellipse, we have:
\[ ae = 1 \] (from the foci \( (\pm 1, 0) \))

Substituting \( a \):
\[ \frac{1}{\sqrt{2}} e = 1 \]
Thus,
\[ e = \sqrt{2} \]

### Step 6: Relate Eccentricity to \( b \)
For a hyperbola, the eccentricity is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values we have:
\[ \sqrt{2} = \sqrt{1 + \frac{b^2}{(\frac{1}{\sqrt{2}})^2}} \]
This simplifies to:
\[ \sqrt{2} = \sqrt{1 + 2b^2} \]

### Step 7: Solve for \( b^2 \)
Squaring both sides:
\[ 2 = 1 + 2b^2 \]
Thus:
\[ 2b^2 = 1 \]
\[ b^2 = \frac{1}{2} \]

### Step 8: 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 \( a^2 = \frac{1}{2} \) and \( b^2 = \frac{1}{2} \):
\[ \frac{x^2}{\frac{1}{2}} - \frac{y^2}{\frac{1}{2}} = 1 \]

This simplifies to:
\[ 2x^2 - 2y^2 = 1 \]

### Final Equation of the Hyperbola
Thus, the equation of the hyperbola is:
\[ 2x^2 - 2y^2 = 1 \]