The ellipse `x^(2)/169+y^(2)/25=1` has the same eccentricity as the ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1`. What is the ratio of a to b ?

Options

A. `5//13`
B. `13//5`
C. `7//8`
D. `8//7`

Explanation

To solve the problem, we need to find the ratio of \( a \) to \( b \) for the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), which has the same eccentricity as the ellipse \( \frac{x^2}{169} + \frac{y^2}{25} = 1 \).

### Step-by-step Solution:

1. **Identify the parameters of the first ellipse**:
The given ellipse is \( \frac{x^2}{169} + \frac{y^2}{25} = 1 \).
Here, \( a^2 = 169 \) and \( b^2 = 25 \).
Thus, \( a = \sqrt{169} = 13 \) and \( b = \sqrt{25} = 5 \).

2. **Calculate the eccentricity of the first ellipse**:
The formula for the eccentricity \( e \) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values:
\[ e = \sqrt{1 - \frac{25}{169}} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{169 - 25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13} \]

3. **Set up the equation for the second ellipse**:
The second ellipse is given by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
We know that its eccentricity is also \( \frac{12}{13} \). Therefore, we can write:
\[ \frac{12}{13} = \sqrt{1 - \frac{b^2}{a^2}} \]

4. **Square both sides to eliminate the square root**:
\[ \left(\frac{12}{13}\right)^2 = 1 - \frac{b^2}{a^2} \]
This simplifies to:
\[ \frac{144}{169} = 1 - \frac{b^2}{a^2} \]

5. **Rearranging the equation**:
\[ \frac{b^2}{a^2} = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169} \]

6. **Taking the square root**:
\[ \frac{b}{a} = \frac{5}{13} \]

7. **Finding the ratio \( \frac{a}{b} \)**:
To find \( \frac{a}{b} \), we take the reciprocal:
\[ \frac{a}{b} = \frac{13}{5} \]

### Conclusion:
The ratio \( \frac{a}{b} \) is \( \frac{13}{5} \).