The equation of the hyperbola whose foci are `(0, pm 13)` and length of conjugate axis is 24 is (i) `144y^(2) - 25 x^(2) = 3600` (ii) `144 x^(2) - 25 y^(2) = 3600` (iii) `25x^(2)- 144 y^(2) = 3600` (iv) `25y^(2)-144x^(2)=3600`

Options

A. `144y^(2) - 25 x^(2) = 3600`
B. `144 x^(2) - 25 y^(2) = 3600`
C. `25x^(2)- 144 y^(2) = 3600`
D. `25y^(2)-144x^(2)=3600`

Answer

To find the equation of the hyperbola given the foci and the length of the conjugate axis, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the foci and conjugate axis**:
The foci of the hyperbola are given as (0, ±13). This means that the hyperbola is vertical and centered at the origin (0,0). The length of the conjugate axis is given as 24.

2. **Determine the values of 'c' and 'b'**:
The distance from the center to each focus is denoted as 'c'. Since the foci are at (0, ±13), we have:
\[ c = 13 \]
The length of the conjugate axis is 24, so the semi-conjugate axis 'b' is:
\[ 2b = 24 \implies b = 12 \]

3. **Use the relationship between 'a', 'b', and 'c'**:
For hyperbolas, the relationship between 'a', 'b', and 'c' is given by:
\[ c^2 = a^2 + b^2 \]
Substituting the known values:
\[ 13^2 = a^2 + 12^2 \]
\[ 169 = a^2 + 144 \]
\[ a^2 = 169 - 144 = 25 \]

4. **Write the standard form of the hyperbola**:
The standard equation of a hyperbola that opens vertically is:
\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]
Substituting the values of \(a^2\) and \(b^2\):
\[ \frac{y^2}{25} - \frac{x^2}{144} = 1 \]

5. **Clear the denominators**:
To express this in a standard form without fractions, multiply through by the least common multiple (LCM) of the denominators (which is 3600):
\[ 144y^2 - 25x^2 = 3600 \]

6. **Final equation**:
Thus, the equation of the hyperbola is:
\[ 144y^2 - 25x^2 = 3600 \]

### Conclusion:
The correct option is (i) \( 144y^2 - 25x^2 = 3600 \).