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

The vertices (y+4)^2-((x-2)^2)/4 =1 is (2,-3),(2,-5)

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Refine

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

- x^{2} + 4 y^{2} + 2 x + 24 y + 31 = 0

9 x^{2} - 4 y^{2} = 144

4 y^{2} - 9 x^{2} + 40 y - 540 x - 8036 = 0

11 x^{2} - 25 y^{2} + 22 x + 250 y - 889 = 0

4 x^{2} - 45 y^{2} = 180

What is the vertices (y+4)^2-((x-2)^2)/4 =1 ?
The vertices (y+4)^2-((x-2)^2)/4 =1 is (2,-3),(2,-5)