Conic Sections Question 384
Question: Equation of the ellipse with eccentricity $ \frac{1}{2} $ and foci at $ (\pm 1,\ 0) $ is
[MP PET 2002]
Options:
A) $ \frac{x^{2}}{3}+\frac{y^{2}}{4}=1 $
B) $ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1 $
C) $ \frac{x^{2}}{3}+\frac{y^{2}}{4}=\frac{4}{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Given that, $ e=\frac{1}{2} $ and $ (\pm ae,0)=(\pm 1,0) $
$ \Rightarrow ae=1 $
$ \Rightarrow a=2 $ . Now $ b^{2}=a^{2}(1-e^{2}) $
$ \Rightarrow b^{2}=4( 1-\frac{1}{4} ) $
$ \Rightarrow b^{2}=3 $
Hence, equation of ellipse is $ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1. $
BETA
