Conic Sections Question 478
Question: Centre of hyperbola $ 9x^{2}-16y^{2}+18x+32y-151=0 $ is
Options:
A) (1, -1)
B) (-1, 1)
C) (-1, -1)
D) (1, 1)
Show Answer
Answer:
Correct Answer: B
Solution:
Centre is given by $ ( \frac{hf-bg}{ab-h^{2}},\frac{gh-af}{ab-h^{2}} )=( \frac{+16.9}{-9.16},\frac{-9(16)}{-9(16)} )=(-1,1) $ .