What is the vertices ((y+4)^2)/9-((x+2)^2)/4 =1 ?

The vertices ((y+4)^2)/9-((x+2)^2)/4 =1 is (-2,-1),(-2,-7)

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vertices ((y+4)^2)/9-((x+2)^2)/4 =1

vertices

\frac{( y + 4 ) ^{2}}{9} - \frac{( x + 2 ) ^{2}}{4} = 1

(- 2, - 1), (- 2, - 7)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 4 ) ^{2}}{9} - \frac{( x + 2 ) ^{2}}{4} = 1 :

Up-down Hyperbola with

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

(- 2, - 4 + 3), (- 2, - 4 - 3)

Refine

(- 2, - 1), (- 2, - 7)

x^{2} - 3 y^{2} - 2 x - 6 y + 7 = 0

\frac{x ^{2}}{29} - \frac{y ^{2}}{20} = 1

y^{2} - x^{2} = \frac{9}{25}

16 y^{2} - 9 x^{2} + 36 x + 96 y - 36 = 0

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

What is the vertices ((y+4)^2)/9-((x+2)^2)/4 =1 ?
The vertices ((y+4)^2)/9-((x+2)^2)/4 =1 is (-2,-1),(-2,-7)