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) $ .