(i) Find the equation of hyperbola whose eccentricity is `sqrt((13)/(12))` and the distance between foci is 26.
(ii) The foci of a hyperbola coincide of the ellipse `9x^(2)+25y^(2)=225`. If the eccentricity of the hyperbola is 2, then find its equation.

Answer

Let's solve the problem step by step.

### Part (i)

**Given:**
- Eccentricity \( e = \sqrt{\frac{13}{12}} \)
- Distance between foci = 26

**Step 1: Find the value of \( a \)**
The distance between the foci of a hyperbola is given by \( 2ae \).
So, we have:
\[ 2ae = 26 \implies ae = 13 \]

**Step 2: Substitute the value of \( e \)**
Substituting the value of \( e \):
\[ a \cdot \sqrt{\frac{13}{12}} = 13 \]

**Step 3: Solve for \( a \)**
Squaring both sides:
\[ a^2 \cdot \frac{13}{12} = 169 \]
\[ a^2 = 169 \cdot \frac{12}{13} = 156 \]

**Step 4: Find \( b^2 \)**
Using the relationship \( b^2 = a^2(e^2 - 1) \):
\[ e^2 = \frac{13}{12} \implies e^2 - 1 = \frac{13}{12} - 1 = \frac{1}{12} \]
Thus,
\[ b^2 = 156 \cdot \frac{1}{12} = 13 \]

**Step 5: 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 \( a^2 = 156 \) and \( b^2 = 13 \):
\[ \frac{x^2}{156} - \frac{y^2}{13} = 1 \]

### Part (ii)

**Given:**
- The foci of the hyperbola coincide with the foci of the ellipse \( 9x^2 + 25y^2 = 225 \)
- Eccentricity of the hyperbola \( e = 2 \)

**Step 1: Write the equation of the ellipse in standard form**
Dividing the equation by 225:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]

**Step 2: Identify \( a \) and \( b \) of the ellipse**
From the standard form:
- \( a^2 = 25 \) (where \( a = 5 \))
- \( b^2 = 9 \) (where \( b = 3 \))

**Step 3: Find the eccentricity of the ellipse**
Using the formula for the eccentricity of the ellipse:
\[ e_{ellipse} = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

**Step 4: Find the foci of the ellipse**
The foci of the ellipse are located at \( \pm ae \):
\[ F_{ellipse} = 5 \cdot \frac{4}{5} = 4 \]

**Step 5: Set the foci of the hyperbola**
Since the foci of the hyperbola coincide with the foci of the ellipse, we have:
\[ F_{hyperbola} = 4 \]
This means:
\[ ae = 4 \]

**Step 6: Solve for \( a \)**
Substituting \( e = 2 \):
\[ a \cdot 2 = 4 \implies a = 2 \]

**Step 7: Find \( b^2 \)**
Using the relationship \( b^2 = a^2(e^2 - 1) \):
\[ e^2 = 4 \implies e^2 - 1 = 3 \]
Thus,
\[ b^2 = 2^2 \cdot 3 = 4 \cdot 3 = 12 \]

**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 \( a^2 = 4 \) and \( b^2 = 12 \):
\[ \frac{x^2}{4} - \frac{y^2}{12} = 1 \]

### Final Answers:
1. The equation of the hyperbola in part (i) is:
\[ \frac{x^2}{156} - \frac{y^2}{13} = 1 \]
2. The equation of the hyperbola in part (ii) is:
\[ \frac{x^2}{4} - \frac{y^2}{12} = 1 \]