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Centre of hyperbola $9 x^2-16 y^2+18 x+32 y-151=0$ is
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Verified Answer
The correct answer is:
$(-1,1)$
Centre is given by
$\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}\right)=\left(\frac{+16.9}{-9.16}, \frac{-9(16)}{-9(16)}\right)=(-1,1)$
$\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}\right)=\left(\frac{+16.9}{-9.16}, \frac{-9(16)}{-9(16)}\right)=(-1,1)$
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