Conic Sections Question 112
Question: The equation of the hyperbola whose foci are (6, 4) and (-4, 4) and eccentricity 2 is given by
[MP PET 1993]
Options:
A) $ 12x^{2}-4y^{2}-24x+32y-127=0 $
B) $ 12x^{2}+4y^{2}+24x-32y-127=0 $
C) $ 12x^{2}-4y^{2}-24x-32y+127=0 $
D) $ 12x^{2}-4y^{2}+24x+32y+127=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Foci are (6,4) and (-4,4), $ e=2 $ and centre is $ ( \frac{6-4}{2},4 )=(1,4) $
therefore $ 6=1+ae $
therefore $ ae=5 $
therefore $ a=\frac{5}{2} $ and $ b=\frac{5}{2}(\sqrt{3}) $
Hence the required equation is $ \frac{{{(x-1)}^{2}}}{(25/4)}-\frac{{{(y-4)}^{2}}}{(75/4)}=1 $
or $ 12x^{2}-4y^{2}-24x+32y-127=0 $ .
BETA
