The eccentricity of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is:

Question

Accepted Answer

For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, the eccentricity $e$ is given by the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$. Here $a^2 = 25$ and $b^2 = 9$. Thus, $e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Conic Sections

❓ Sibling Questions in this Chapter