What is the vertices 16y^2-32y-9x^2-18x=137 ?

The vertices 16y^2-32y-9x^2-18x=137 is (-1,4),(-1,-2)

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vertices 16y^2-32y-9x^2-18x=137

vertices

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

(- 1, 4), (- 1, - 2)

Solution steps

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

Calculate Hyperbola properties

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137 :

Up-down Hyperbola with

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

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

Refine

(- 1, 4), (- 1, - 2)

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

\frac{y ^{2}}{64} - \frac{x ^{2}}{36} = 1

4 x^{2} - y^{2} - 24 x - 4 y + 28 = 0

\frac{y ^{2}}{1} - \frac{x ^{2}}{15} = 1

4 y^{2} - 49 x^{2} - 392 x + 16 y - 964 = 0

What is the vertices 16y^2-32y-9x^2-18x=137 ?
The vertices 16y^2-32y-9x^2-18x=137 is (-1,4),(-1,-2)