Conic Sections Question 427
Question: The foci of the hyperbola $ 2x^{2}-3y^{2}=5 $ , is
[MP PET 2000]
Options:
A) $ ( \pm \frac{5}{\sqrt{6}},\ 0 ) $
B) $ ( \pm \frac{5}{6},\ 0 ) $
C) $ ( \pm \frac{\sqrt{5}}{6},\ 0 ) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
The given equation is $ 2x^{2}-3y^{2}=5 $
$ \Rightarrow $ $ \frac{x^{2}}{5/2}-\frac{y^{2}}{5/3}=1 $
Now $ b^{2}=a^{2}(e^{2}-1) $
$ \Rightarrow \frac{5}{3}=\frac{5}{2}(e^{2}-1) $
therefore $ e=\sqrt{\frac{5}{3}} $ . The foci of hyperbola $ (\pm ae,0) $
$ =( \pm \sqrt{\frac{5}{2}}.\sqrt{\frac{5}{3}},0 ) $
$ =( \pm \frac{5}{\sqrt{6}},0 ) $ .
BETA
