The vertices of the hyperbola $9 x^{2} - 16 y^{2} - 36 x + 96 y - 252 = 0$ are:

(6,3)

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(-6,3)

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(2,3)

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(-2,3)

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Solution

The correct option is D (-2,3)
$T h e e q u a t i o n c a n b e r e w r i t t e n a s :$
$9 (x^{2} - 4 x + 4 - 4) - 16 (y^{2} - 6 y + 9 - 9) = 252 9 {(x - 2)}^{2} - 16 {(y - 3)}^{2} = 252 + 36 - 144 or \frac{{(x - 2)}^{2}}{16} - \frac{{(y - 3)}^{2}}{9} = 1 or \frac{X^{2}}{A^{2}} - \frac{Y^{2}}{B^{2}} = 1 Hence the vertices are : X = \pm A, Y = 0 or x - 2 = \pm 4, y - 3 = 0 or x = 6, - 2 and y = 3 Hence, the vertices are : (6, 3), (- 2, 3)$

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