Conic Sections Question 432
Question: The foci of the hyperbola $ 9x^{2}-16y^{2}=144 $ are
[MP PET 2001]
Options:
A) $ (\pm 4,\ 0) $
B) $ (0,\ \pm 4) $
C) $ (\pm 5,\ 0) $
D) $ (0,\ \pm 5) $
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Answer:
Correct Answer: C
Solution:
The equation of hyperbola is $ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 $ Now $ b^{2}=a^{2}(e^{2}-1) $
$ \Rightarrow e=\frac{5}{4} $
Hence foci are $ (\pm ae,0) $
therefore $ ( \pm 4.\frac{5}{4},0 ) $ i.e., $ (\pm 5,0) $ .
BETA
