The vertices of the hyperbola 9x2-16y2-36x-96y-252-0 nbsp-are-6-3-6-3-2-3-2-3-

The correct options are A (6,3) D ( 2,3)The nbsp;equation nbsp;can nbsp;be nbsp;rewritten nbsp;as: 9x2 4x+4 4 16y2 6y+9 9=2529x 22 16y 32=252+36 144or ...

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Byju's Answer

Standard XII

Mathematics

Vertices of Hyperbola

The vertices ...

Question

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 options are
A (6,3)
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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Similar questions

Q. Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is
(a) 16x² − 9y² = 144
(b) 9x² − 16y² = 144
(c) 25x² − 9y² = 225
(d) 9x² − 25y² = 81

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9 x^{2} - 16 y^{2} + 18 x + 32 y - 151 = 0

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Q. Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is
(a) 16x² − 9y² = 144
(b) 9x² − 16y² = 144
(c) 25x² − 9y² = 225
(d) 9x² − 25y² = 81

Q. The foci of the hyperbola

9 x^{2} - 16 y^{2} - 18 x + 32 y - 151 = 0

are

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The vertices of the hyperbola 9x2−16y2−36x+96y−252=0 are:

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