Conic Sections Question 285
Question: The distance between the foci of the ellipse $ 3x^{2}+4y^{2}=48 $ is
Options:
A) 2
B) 4
C) 6
D) 8
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{x^{2}}{(48/3)}+\frac{y^{2}}{(48/4)}=1 $
$ a^{2}=16,\ b^{2}=12 $
therefore $ e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\frac{1}{2} $
Distance is $ 2ae=2\cdot 4\cdot \frac{1}{2}=4 $ .
BETA
