If the locus of the point which moves so that the difference (p) 0 of its distance from the points `(5, 0) and (-5,0)` is 2 is `x^2/a^2-y^2/24=1` then a is

Answer

To solve the problem, we need to find the value of \( a \) in the equation of the hyperbola given by:

\[ \frac{x^2}{a^2} - \frac{y^2}{24} = 1 \]

where the locus of the point is defined by the difference of its distances from the points \( (5, 0) \) and \( (-5, 0) \) being equal to 2.

### Step-by-Step Solution:

1. **Identify the Foci of the Hyperbola:**
The foci of the hyperbola are located at \( (c, 0) \) and \( (-c, 0) \), where \( c = \sqrt{a^2 + b^2} \). Here, the foci are \( (5, 0) \) and \( (-5, 0) \). Therefore, we have:
\[ c = 5 \]

2. **Identify the Value of \( b^2 \):**
From the hyperbola equation, we know that \( b^2 = 24 \).

3. **Use the Relationship Between \( a \), \( b \), and \( c \):**
The relationship between \( a \), \( b \), and \( c \) for a hyperbola is given by:
\[ c^2 = a^2 + b^2 \]
Substituting the known values:
\[ 5^2 = a^2 + 24 \]
This simplifies to:
\[ 25 = a^2 + 24 \]

4. **Solve for \( a^2 \):**
Rearranging the equation gives:
\[ a^2 = 25 - 24 = 1 \]

5. **Find \( a \):**
Taking the square root of both sides:
\[ a = \sqrt{1} = 1 \]

### Conclusion:
Thus, the value of \( a \) is:

\[ \boxed{1} \]