Conic Sections Question 448
Question: The equation of the hyperbola whose foci are the foci of the ellipse $ \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 $ and the eccentricity is 2, is
Options:
A) $ \frac{x^{2}}{4}+\frac{y^{2}}{12}=1 $
B) $ \frac{x^{2}}{4}-\frac{y^{2}}{12}=1 $
C) $ \frac{x^{2}}{12}+\frac{y^{2}}{4}=1 $
D) $ \frac{x^{2}}{12}-\frac{y^{2}}{4}=1 $
Show Answer
Answer:
Correct Answer: B
Solution:
Here for given ellipse $ a=5,\ b=3,\ b^{2}=a^{2}(1-e^{2}) $
therefore $ e=\frac{4}{5} $
Therefore, focus is (- 4, 0), (4, 0). Given eccentricity of hyperbola = 2 $ a=\frac{ae}{e}=\frac{4}{2}=2 $ and $ b=2\sqrt{(4-1)}=2\sqrt{3} $
Hence hyperbola is $ \frac{x^{2}}{4}-\frac{y^{2}}{12}=1 $ .
BETA
