Conic Sections Question 475
Question: The eccentricity of the hyperbola $ \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 $ is
[Karnataka CET 2005]
Options:
A) 3/4
B) 3/5
C) $ \sqrt{41}/4 $
D) $ \sqrt{41/5} $
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Answer:
Correct Answer: C
Solution:
Equation of hyperbola is $ \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 $
Eccentricity is $ e^{2}=\frac{b^{2}}{a^{2}}+1 $ i.e., $ e^{2}=\frac{25}{16}+1 $
therefore $ e^{2}=\frac{41}{16} $
therefore $ e=\frac{\sqrt{41}}{4} $ .
BETA
