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

The vertices ((y+2)^2)/2-((x+1)^2)/2 =1 is (-1,-2+sqrt(2)),(-1,-2-sqrt(2))

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

\frac{y ^{2}}{16} - \frac{x ^{2}}{4} = 1

x^{2} - y^{2} - 4 x + 4 y + 9 = 0

x^{2} - y^{2} + 8 x - 2 y - 10 = 0

25 y^{2} - 9 x^{2} = 225

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