`E_1:x^2/81 +y^2/b^2=1` and `E_2: x^2/b^2+y^2/49=1` two ellipses having the same eccentricity if `b^2` is equal to

Options

A. 32
B. 63
C. 65
D. 64

Answer

To solve the problem, we need to find the value of \( b^2 \) such that the two given ellipses have the same eccentricity.

### Step-by-Step Solution:

1. **Identify the given ellipses**:
- The first ellipse \( E_1 \) is given by the equation:
\[ \frac{x^2}{81} + \frac{y^2}{b^2} = 1 \]
- The second ellipse \( E_2 \) is given by the equation:
\[ \frac{x^2}{b^2} + \frac{y^2}{49} = 1 \]

2. **Determine the eccentricity formula**:
The eccentricity \( e \) of an ellipse in the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
where \( a^2 \) is the larger denominator (semi-major axis squared) and \( b^2 \) is the smaller denominator (semi-minor axis squared).

3. **Calculate eccentricity for both ellipses**:
- For \( E_1 \):
- Here, \( a^2 = 81 \) and \( b^2 = b^2 \).
- Therefore, the eccentricity \( e_1 \) is:
\[ e_1 = \sqrt{1 - \frac{b^2}{81}} \]

- For \( E_2 \):
- Here, \( a^2 = b^2 \) and \( b^2 = 49 \).
- Therefore, the eccentricity \( e_2 \) is:
\[ e_2 = \sqrt{1 - \frac{49}{b^2}} \]

4. **Set the eccentricities equal**:
Since the problem states that the two ellipses have the same eccentricity, we set \( e_1 = e_2 \):
\[ \sqrt{1 - \frac{b^2}{81}} = \sqrt{1 - \frac{49}{b^2}} \]

5. **Square both sides to eliminate the square roots**:
\[ 1 - \frac{b^2}{81} = 1 - \frac{49}{b^2} \]
Simplifying this gives:
\[ -\frac{b^2}{81} = -\frac{49}{b^2} \]
or
\[ \frac{b^2}{81} = \frac{49}{b^2} \]

6. **Cross-multiply to solve for \( b^4 \)**:
\[ b^4 = 49 \times 81 \]

7. **Calculate \( 49 \times 81 \)**:
\[ 49 \times 81 = 3969 \]
Thus, we have:
\[ b^4 = 3969 \]

8. **Find \( b^2 \)**:
Taking the square root of both sides:
\[ b^2 = \sqrt{3969} = 63 \]

### Final Answer:
The value of \( b^2 \) is:
\[ \boxed{63} \]