Conic Sections Question 365
Question: Eccentricity of the ellipse $ 9x^{2}+25y^{2}=225 $ is
[Kerala (Engg.) 2002]
Options:
A) $ \frac{3}{5} $
B) $ \frac{4}{5} $
C) $ \frac{9}{25} $
D) $ \frac{\sqrt{34}}{5} $
Show Answer
Answer:
Correct Answer: B
Solution:
The ellipse is $ \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 $
$ \therefore e=\sqrt{1-\frac{b^{2}}{a^{2}}} $ i.e., $ \sqrt{1-\frac{9}{25}}=\frac{4}{5} $ .
BETA
