Question

If the eccentricities 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 $\frac{a}{b}$, is

• A. $\frac{5}{11}$
• B. $\frac{6}{13}$
• C. $\frac{13}{5}$
• D. $\frac{13}{6}$

Answer 1

The correct option is C $\frac{13}{5}$

Given equation of ellipse are

$\frac{x^2}{169}+\frac{y^2}{25}=1\to \left(1\right)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\to \left(2\right)$

Eccentricity of ellipse is given by

$e^2=1-\frac{b^2}{a^2}$

From $\left(1\right)$

$a^2=169b^2=25e^2=1-\frac{25}{169}=\frac{169-25}{169}=\frac{144}{169}$

So As eccentricities is equal

So

$\frac{144}{169}=1-\frac{b^2}{a^2}\frac{b^2}{a^2}=1-\frac{144}{169}=\frac{169-144}{169}=\frac{25}{169}\Rightarrow \left(\frac{b}{a}\right)=\frac{5}{13}\therefore \frac{a}{b}=\frac{13}{5}$