The foci of an ellipse coincide with foci of the hyperbola `3x^(2)-y^(2)=12`. Find the equation of the ellipse, if its eccentricity is `(4)/(5)`.

Answer

To solve the problem step by step, we will find the equation of the ellipse whose foci coincide with the foci of the given hyperbola and has a specified eccentricity.

### Step 1: Write the equation of the hyperbola
The given hyperbola is:
\[ 3x^2 - y^2 = 12 \]

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

### Step 3: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:
- \(a^2 = 4\)
- \(b^2 = 12\)

### Step 4: Calculate the eccentricity of the hyperbola
The eccentricity \(e\) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ e = \sqrt{1 + \frac{12}{4}} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 5: Calculate the foci of the hyperbola
The foci of the hyperbola are located at:
\[ \text{Foci} = (\pm ae, 0) \]
Where \(a = \sqrt{a^2} = \sqrt{4} = 2\). Thus:
\[ \text{Foci} = (\pm 2 \cdot 2, 0) = (\pm 4, 0) \]

### Step 6: Set up the ellipse parameters
The eccentricity of the ellipse is given as \( \frac{4}{5} \). Let \(a'\) and \(b'\) be the semi-major and semi-minor axes of the ellipse, respectively. The foci of the ellipse are also at \((\pm a'e', 0)\) where \(e' = \frac{4}{5}\).

### Step 7: Calculate the foci of the ellipse
The foci of the ellipse are:
\[ \text{Foci} = (a' \cdot e', 0) = (a' \cdot \frac{4}{5}, 0) \]
Setting this equal to the foci of the hyperbola:
\[ a' \cdot \frac{4}{5} = 4 \]
Solving for \(a'\):
\[ a' = 4 \cdot \frac{5}{4} = 5 \]

### Step 8: Calculate \(a'^2\)
Now, we find \(a'^2\):
\[ a'^2 = 5^2 = 25 \]

### Step 9: Calculate \(b'^2\)
Using the relationship for the ellipse:
\[ b'^2 = a'^2(1 - e'^2) \]
First, calculate \(e'^2\):
\[ e'^2 = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]
Now substitute into the equation for \(b'^2\):
\[ b'^2 = 25 \left(1 - \frac{16}{25}\right) = 25 \left(\frac{9}{25}\right) = 9 \]

### Step 10: Write the equation of the ellipse
The equation of the ellipse in standard form is:
\[ \frac{x^2}{a'^2} + \frac{y^2}{b'^2} = 1 \]
Substituting the values of \(a'^2\) and \(b'^2\):
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]

### Final Answer
Thus, the equation of the ellipse is:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]