Conic Sections Question 411
Question: If $ (0,\ \pm 4) $ and $ (0,\ \pm 2) $ be the foci and vertices of a hyperbola, then its equation is
Options:
A) $ \frac{x^{2}}{4}-\frac{y^{2}}{12}=1 $
B) $ \frac{x^{2}}{12}-\frac{y^{2}}{4}=1 $
C) $ \frac{y^{2}}{4}-\frac{x^{2}}{12}=1 $
D) $ \frac{y^{2}}{12}-\frac{x^{2}}{4}=1 $
Show Answer
Answer:
Correct Answer: C
Solution:
Foci $ (0,\pm 4) $
$ \equiv (0,\pm be) $
therefore $ be=4 $ Vertices $ (0,\pm 2)\equiv (0,\pm b)\Rightarrow b=2\Rightarrow a=2\sqrt{3} $
Hence equation is $ \frac{-x^{2}}{{{(2\sqrt{3})}^{2}}}+\frac{y^{2}}{{{(2)}^{2}}}=1 $ or $ \frac{y^{2}}{4}-\frac{x^{2}}{12}=1 $ .
BETA
