Conic Sections — Easy JEE math MCQ
The eccentricity of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is:
- A. $\frac{3}{5}$
- B. $\frac{4}{5}$
- C. $\frac{5}{4}$
- D. $\frac{5}{3}$
Solution
$a^2 = 25, b^2 = 16 \Rightarrow a = 5, b = 4$
$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$