The centre of the hyperbola 9x^2 - 16y^2 + 18 | vedclass

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(B) The given equation of the hyperbola is $9x^2 - 16y^2 + 18x + 32y - 151 = 0$. Rearranging the terms,we get: $9(x^2 + 2x) - 16(y^2 - 2y) = 151$. Com

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$(-1, 1)$

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(B) The given equation of the hyperbola is $9x^2 - 16y^2 + 18x + 32y - 151 = 0$. Rearranging the terms,we get: $9(x^2 + 2x) - 16(y^2 - 2y) = 151$. Completing the square for $x$ and $y$: $9(x^2 + 2x + 1) - 16(y^2 - 2y + 1) = 151 + 9 - 16$. $9(x + 1)^2 - 16(y - 1)^2 = 144$. Dividing by $144$: $\frac{(x + 1)^2}{16} - \frac{(y - 1)^2}{9} = 1$. Comparing this with the standard form $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$,the centre $(h, k)$ is $(-1, 1)$.

The centre of the hyperbola $9x^2 - 16y^2 + 18x + 32y - 151 = 0$ is

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