What is the vertices 9x^2-16y^2+54x-16y=-113 ?

The vertices 9x^2-16y^2+54x-16y=-113 is (-3,1),(-3,-2)

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vertices 9x^2-16y^2+54x-16y=-113

vertices

9 x^{2} - 16 y^{2} + 54 x - 16 y = - 113

(- 3, 1), (- 3, - 2)

Solution steps

(h, k + a), (h, k - a)

Calculate Hyperbola properties

9 x^{2} - 16 y^{2} + 54 x - 16 y = - 113 :

Up-down Hyperbola with

(h, k) = (- 3, - \frac{1}{2}), a = \frac{3}{2}, b = 2

(- 3, - \frac{1}{2} + \frac{3}{2}), (- 3, - \frac{1}{2} - \frac{3}{2})

Refine

(- 3, 1), (- 3, - 2)

(\frac{y - 4}{4})^{2} - (\frac{x - 6}{4})^{2} = 1

4 x^{2} - 9 y^{2} - 16 x - 72 y - 129 = 0

\frac{x ^{2}}{5} - \frac{y ^{2}}{7} = 1

- \frac{x ^{2}}{144} + \frac{y ^{2}}{25} = 1

16 x^{2} - 9 y^{2} - 64 x + 54 y - 161 = 0

What is the vertices 9x^2-16y^2+54x-16y=-113 ?
The vertices 9x^2-16y^2+54x-16y=-113 is (-3,1),(-3,-2)