What is the vertices 9y^2+36y-4x^2-32x-64=0 ?

The vertices 9y^2+36y-4x^2-32x-64=0 is (-4,0),(-4,-4)

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vertices 9y^2+36y-4x^2-32x-64=0

vertices

9 y^{2} + 36 y - 4 x^{2} - 32 x - 64 = 0

(- 4, 0), (- 4, - 4)

Solution steps

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

Calculate Hyperbola properties

9 y^{2} + 36 y - 4 x^{2} - 32 x - 64 = 0 :

Up-down Hyperbola with

(h, k) = (- 4, - 2), a = 2, b = 3

(- 4, - 2 + 2), (- 4, - 2 - 2)

Refine

(- 4, 0), (- 4, - 4)

- 25 x^{2} + 9 y^{2} = 225

\frac{( y - 2 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{25} = 1

9 y^{2} - x^{2} + 54 y + 2 x + 62 = 0

x^{2} - y^{2} = 6

4 x^{2} - y^{2} + 24 x - 4 y = 4

What is the vertices 9y^2+36y-4x^2-32x-64=0 ?
The vertices 9y^2+36y-4x^2-32x-64=0 is (-4,0),(-4,-4)