What is the vertices-9x^2+16y^2-90x+128y-113=0 ?

The vertices-9x^2+16y^2-90x+128y-113=0 is (-5,-1),(-5,-7)

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vertices-9x^2+16y^2-90x+128y-113=0

vertices

- 9 x^{2} + 16 y^{2} - 90 x + 128 y - 113 = 0

(- 5, - 1), (- 5, - 7)

Solution steps

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

Calculate Hyperbola properties

- 9 x^{2} + 16 y^{2} - 90 x + 128 y - 113 = 0 :

Up-down Hyperbola with

(h, k) = (- 5, - 4), a = 3, b = 4

(- 5, - 4 + 3), (- 5, - 4 - 3)

Refine

(- 5, - 1), (- 5, - 7)

2 y^{2} - 3 x^{2} - 4 y + 12 x + 8 = 0

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{18} = 1

\frac{x ^{2}}{1} - \frac{y ^{2}}{35} = 1

\frac{( x + 6 ) ^{2}}{9} = \frac{( y - 2 ) ^{2}}{16} + 1

\frac{x ^{2}}{27} - \frac{y ^{2}}{72} = 1

What is the vertices-9x^2+16y^2-90x+128y-113=0 ?
The vertices-9x^2+16y^2-90x+128y-113=0 is (-5,-1),(-5,-7)