What is the vertices 64x^2+128x-9y^2-72y-656=0 ?

The vertices 64x^2+128x-9y^2-72y-656=0 is (2,-4),(-4,-4)

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vertices 64x^2+128x-9y^2-72y-656=0

vertices

64 x^{2} + 128 x - 9 y^{2} - 72 y - 656 = 0

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

Solution steps

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

Calculate Hyperbola properties

64 x^{2} + 128 x - 9 y^{2} - 72 y - 656 = 0 :

Right-left Hyperbola with

(h, k) = (- 1, - 4), a = 3, b = 8

(- 1 + 3, - 4), (- 1 - 3, - 4)

Refine

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

\frac{x ^{2}}{4} - \frac{y ^{2}}{12} = 1

9 y^{2} - 4 x^{2} - 36 y - 8 x = 4

4 x^{2} - 25 y^{2} = 100

\frac{( x - 1 ) ^{2}}{9} - \frac{( y - 2 ) ^{2}}{16} = 1

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

What is the vertices 64x^2+128x-9y^2-72y-656=0 ?
The vertices 64x^2+128x-9y^2-72y-656=0 is (2,-4),(-4,-4)