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

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

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Refine

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

- 16 x^{2} + 9 y^{2} = 144

x^{2} - y^{2} + 4 x - 6 y = 6

\frac{x ^{2}}{25} - \frac{y ^{2}}{4} = 1

x^{2} - 4 y^{2} + 2 x + 8 y - 7 = 0

4 x^{2} - 25 y^{2} - 40 x - 150 y - 225 = 0

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