Conic Sections Question 79
Question: For the ellipse $ 25x^{2}+9y^{2}-150x-90y+225=0 $ the eccentricity $ e= $
[Karnataka CET 2004]
Options:
A) 2/5
B) 3/5
C) 4/5
D) 1/5
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation of ellipse is, $ 25x^{2}+9y^{2}-150x-90y+225=0 $
$ \Rightarrow $ $ 25{{(x-3)}^{2}}+9{{(y-5)}^{2}}=225 $
$ \Rightarrow $ $ \frac{{{(x-3)}^{2}}}{9}+\frac{{{(y-5)}^{2}}}{25} $ = 1. Here $ b>a $
$ \therefore $ Eccentricity $ e=\sqrt{1-\frac{a^{2}}{b^{2}}}=\sqrt{1-\frac{9}{25}} $
$ =\sqrt{\frac{16}{25}}=\frac{4}{5} $ .
BETA
