If the eccentricities of the two ellipses fra | vedclass

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(C) For the first ellipse $\frac{x^2}{169} + \frac{y^2}{25} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b_1^2}{a_1^2}} = \sqrt{1 - \fra

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$13/5$

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(C) For the first ellipse $\frac{x^2}{169} + \frac{y^2}{25} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b_1^2}{a_1^2}} = \sqrt{1 - \frac{25}{169}}$.For the second ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the eccentricity $e'$ is $e' = \sqrt{1 - \frac{b^2}{a^2}}$.Given that the eccentricities are equal,$e = e'$,so $\sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{25}{169}}$.Squaring both sides,$1 - \frac{b^2}{a^2} = 1 - \frac{25}{169}$,which implies $\frac{b^2}{a^2} = \frac{25}{169}$.Taking the square root,$\frac{b}{a} = \frac{5}{13}$.Therefore,$\frac{a}{b} = \frac{13}{5}$.

If the eccentricities of the two ellipses $\frac{x^2}{169} + \frac{y^2}{25} = 1$ 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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