Conic Sections Question 476
Question: The equation to the hyperbola having its eccentricity 2 and the distance between its foci is 8
[Karnataka CET 2005]
Options:
A) $ \frac{x^{2}}{12}-\frac{y^{2}}{4}=1 $
B) $ \frac{x^{2}}{4}-\frac{y^{2}}{12}=1 $
C) $ \frac{x^{2}}{8}-\frac{y^{2}}{2}=1 $
D) $ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 $
Show Answer
Answer:
Correct Answer: B
Solution:
Distance between foci = 8 $ 2ae=8 $ also $ e=2 $ ; $ 2a=4 $
therefore $ a=2 $
therefore $ a^{2}=4 $ ;$ b^{2}=4(4-1)=12 $
Equation of hyperbola is $ \frac{x^{2}}{4}-\frac{y^{2}}{12}=1 $ .
BETA
