Conic Sections Question 409
Question: The equation of the transverse and conjugate axis of the hyperbola $ 16x^{2}-y^{2}+64x+4y+44=0 $ are
Options:
A) $ x=2,\ y+2=0 $
B) $ x=2,\ y=2 $
C) $ y=2,\ x+2=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{(4x+8)}^{2}}-{{(y-2)}^{2}}=-44+64-4 $
therefore $ \frac{16{{(x+2)}^{2}}}{16}-\frac{{{(y-2)}^{2}}}{16}=1 $ Transverse and conjugate axes are $ y=2 $ , $ x=-2 $ .
BETA
