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If the eccentricity of the two ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$ and $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are equal, then the value of $a / b$ is
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$13 / 5$
In the first case, eccentricity $e=\sqrt{1-(25 / 169)}$
In the second case, $e^{\prime}=\sqrt{1-\left(b^2 / a^2\right)}$ According to the given condition,
$\begin{aligned} & \sqrt{1-b^2 / a^2}=\sqrt{1-(25 / 169)} \\ & \Rightarrow b / a=5 / 13, \quad(\because a\gt0, b\gt0) \\ & \Rightarrow a / b=13 / 5 .\end{aligned}$
In the second case, $e^{\prime}=\sqrt{1-\left(b^2 / a^2\right)}$ According to the given condition,
$\begin{aligned} & \sqrt{1-b^2 / a^2}=\sqrt{1-(25 / 169)} \\ & \Rightarrow b / a=5 / 13, \quad(\because a\gt0, b\gt0) \\ & \Rightarrow a / b=13 / 5 .\end{aligned}$
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