The length of the latus rectum of the hyperbo | vedclass

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(D) Given equation: $9x^2 - 16y^2 - 18x - 32y - 151 = 0$Rearranging terms: $9(x^2 - 2x) - 16(y^2 + 2y) = 151$Completing the square: $9(x^2 - 2x + 1) -

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$\frac{9}{2}$

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(D) Given equation: $9x^2 - 16y^2 - 18x - 32y - 151 = 0$Rearranging terms: $9(x^2 - 2x) - 16(y^2 + 2y) = 151$Completing the square: $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 with standard form $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$,we get $a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.The length of the latus rectum is given by $\frac{2b^2}{a} = \frac{2 	imes 9}{4} = \frac{18}{4} = \frac{9}{2}$.

The length of the latus rectum of the hyperbola $9x^2 - 16y^2 - 18x - 32y - 151 = 0$ is

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