If the eccentricity of the ellipse `x^2/25+y^2/a^2=1` and `x^2/a^2+y^2/16=1` is same, then the value of `a^2` is :

Options

A. 9
B. 41
C. 15
D. 20

Answer

To solve the problem, we need to find the value of \( a^2 \) given that the eccentricities of the two ellipses are the same. The equations of the ellipses are:

1. \( \frac{x^2}{25} + \frac{y^2}{a^2} = 1 \)
2. \( \frac{x^2}{a^2} + \frac{y^2}{16} = 1 \)

### Step 1: Identify the parameters of the ellipses
For the first ellipse, we have:
- \( a^2 = 25 \)
- \( b^2 = a^2 \)

For the second ellipse, we have:
- \( a^2 = 16 \)
- \( b^2 = a^2 \)

### Step 2: Calculate the eccentricity of the first ellipse
The eccentricity \( e_1 \) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
For the first ellipse:
- \( a^2 = 25 \)
- \( b^2 = a^2 \)

Thus:
\[ e_1 = \sqrt{1 - \frac{a^2}{25}} = \sqrt{1 - \frac{a^2}{25}} \]

### Step 3: Calculate the eccentricity of the second ellipse
For the second ellipse:
- \( a^2 = 16 \)
- \( b^2 = a^2 \)

Thus:
\[ e_2 = \sqrt{1 - \frac{16}{a^2}} \]

### Step 4: Set the eccentricities equal
Since the eccentricities are the same, we set \( e_1 = e_2 \):
\[ \sqrt{1 - \frac{a^2}{25}} = \sqrt{1 - \frac{16}{a^2}} \]

### Step 5: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ 1 - \frac{a^2}{25} = 1 - \frac{16}{a^2} \]

### Step 6: Simplify the equation
Cancelling the 1's from both sides:
\[ -\frac{a^2}{25} = -\frac{16}{a^2} \]
Multiplying through by -1:
\[ \frac{a^2}{25} = \frac{16}{a^2} \]

### Step 7: Cross-multiply to solve for \( a^2 \)
Cross-multiplying gives:
\[ a^4 = 16 \times 25 \]
Calculating the right side:
\[ a^4 = 400 \]

### Step 8: Take the square root
Taking the square root of both sides:
\[ a^2 = \sqrt{400} = 20 \]

### Final Answer
Thus, the value of \( a^2 \) is:
\[ \boxed{20} \]