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 ration of a to b?

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the parameters of the first ellipse**:
The equation of the first ellipse is:
\[ \frac{x^2}{169} + \frac{y^2}{25} = 1 \]
Here, we can identify \( a^2 = 169 \) and \( b^2 = 25 \).

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:
\[ e = \sqrt{1 - \frac{25}{169}} \]
To simplify this, we find a common denominator:
\[ e = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{169 - 25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13} \]

3. **Set up the equation for the second ellipse**:
The second ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Its eccentricity is also given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Since both ellipses have the same eccentricity, we set:
\[ \sqrt{1 - \frac{b^2}{a^2}} = \frac{12}{13} \]

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

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:
\[ \frac{a}{b} = \frac{13}{5} \]

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