The centre of the hyperbola x2 - 4x - 425 - y2 - 6x - 916 - 1 is-

The correct option is C ( 2, 3) the equation of hyperbola is x ^ 2 +4x+4 25 y ^ 2 6x+9 16 =1 (x+2) ^ 2 25 (y 3) ^ 2 16 =1 Therefore ...

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The centre of the hyperbola $\frac{x^{2} + 4 x + 4}{25} - \frac{y^{2} - 6 x + 9}{16} = 1$ is:

(- 4, - 9)

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(- 2, 3)

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(2, - 3)

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(5, 4)

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(25, 16)

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\frac{- x^{2}}{9} + \frac{y^{2}}{16} = 1

\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1

\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1

Equation of the hyperbola with length of the latusrectum $\frac{9}{2}$ and e = $\frac{5}{4}$ is

\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1

\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1

\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1

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