What is the vertices 9x^2-18x-16y^2-64y=199 ?

The vertices 9x^2-18x-16y^2-64y=199 is (5,-2),(-3,-2)

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vertices 9x^2-18x-16y^2-64y=199

vertices

9 x^{2} - 18 x - 16 y^{2} - 64 y = 199

(5, - 2), (- 3, - 2)

Solution steps

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

Calculate Hyperbola properties

9 x^{2} - 18 x - 16 y^{2} - 64 y = 199 :

Right-left Hyperbola with

(h, k) = (1, - 2), a = 4, b = 3

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

Refine

(5, - 2), (- 3, - 2)

\frac{y ^{2}}{49} - \frac{x ^{2}}{16} = 1

- 3 x^{2} + 2 y^{2} + 6 x - 7 y = 7

\frac{( y - 3 ) ^{2}}{49} - \frac{( x + 2 ) ^{2}}{16} = 1

x^{2} - y^{2} - 4 x + 8 y - 21 = 0

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

What is the vertices 9x^2-18x-16y^2-64y=199 ?
The vertices 9x^2-18x-16y^2-64y=199 is (5,-2),(-3,-2)