Conic Sections Question 373
Question: What is the equation of the ellipse with foci $ (\pm 2,\ 0) $ and eccentricity $ =\frac{1}{2} $
[DCE 1999]
Options:
A) $ 3x^{2}+4y^{2}=48 $
B) $ 4x^{2}+3y^{2}=48 $
C) $ 3x^{2}+4y^{2}=0 $
D) $ 4x^{2}+3y^{2}=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Since, $ ae=\pm 2 $
$ \Rightarrow a=\pm 4 $
$ (\because e=1/2) $ Now $ b^{2}=a^{2}(1-e^{2}) $
$ \Rightarrow $ $ b^{2}=16(1-1/4) $
$ \Rightarrow $ $ b^{2}=12 $
Hence ellipse is $ \frac{x^{2}}{16}+\frac{y^{2}}{12}=1 $
$ \Rightarrow $ $ 3x^{2}+4y^{2}=48 $ .
BETA
