If the foci of the ellips ` (x^(2))/( 25)+ ( y^(2))/( 16) =1 `and the hyperbola ` (x^(2))/(4) - ( y^(2))/( b^(2) ) =1` coincide ,then `b^(2)=`

Options

A. a) 4
B. b) 5
C. c) 8
D. d) 9

Answer

To solve the problem, we need to find the value of \( b^2 \) given that the foci of the ellipse and the hyperbola coincide.

### Step-by-Step Solution:

1. **Identify the parameters of the ellipse:**
The equation of the ellipse is given as:
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]
Here, \( a^2 = 25 \) and \( b^2 = 16 \). Thus, we have:
\[ a = \sqrt{25} = 5 \]

2. **Calculate the foci of the ellipse:**
The foci of an ellipse are given by the coordinates \( (\pm c, 0) \), where \( c = \sqrt{a^2 - b^2} \).
\[ c = \sqrt{25 - 16} = \sqrt{9} = 3 \]
Therefore, the foci of the ellipse are at:
\[ (\pm 3, 0) \]

3. **Identify the parameters of the hyperbola:**
The equation of the hyperbola is given as:
\[ \frac{x^2}{4} - \frac{y^2}{b^2} = 1 \]
Here, \( a^2 = 4 \) and we need to find \( b^2 \).

4. **Calculate the foci of the hyperbola:**
The foci of a hyperbola are given by the coordinates \( (\pm c, 0) \), where \( c = \sqrt{a^2 + b^2} \).
\[ c = \sqrt{4 + b^2} \]

5. **Set the foci of the ellipse and hyperbola equal:**
Since the foci of the ellipse and hyperbola coincide, we have:
\[ \sqrt{4 + b^2} = 3 \]

6. **Square both sides to eliminate the square root:**
\[ 4 + b^2 = 9 \]

7. **Solve for \( b^2 \):**
\[ b^2 = 9 - 4 = 5 \]

### Final Answer:
Thus, the value of \( b^2 \) is:
\[ \boxed{5} \]