What is the vertices ((y-5)^2)/4-((x+3)^2)/9 =1 ?

The vertices ((y-5)^2)/4-((x+3)^2)/9 =1 is (-3,7),(-3,3)

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vertices ((y-5)^2)/4-((x+3)^2)/9 =1

vertices

\frac{( y - 5 ) ^{2}}{4} - \frac{( x + 3 ) ^{2}}{9} = 1

(- 3, 7), (- 3, 3)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 5 ) ^{2}}{4} - \frac{( x + 3 ) ^{2}}{9} = 1 :

Up-down Hyperbola with

(h, k) = (- 3, 5), a = 2, b = 3

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

Refine

(- 3, 7), (- 3, 3)

9 x^{2} - 4 y^{2} + 54 x + 16 y + 66 = 0

\frac{y ^{2}}{36} - \frac{x ^{2}}{9} = 1

y^{2} - x^{2} = \frac{1}{25}

\frac{( y + 4 ) ^{2}}{4} - \frac{( x + 4 ) ^{2}}{25} = 1

4 x^{2} - 5 y^{2} + 20 = 0

What is the vertices ((y-5)^2)/4-((x+3)^2)/9 =1 ?
The vertices ((y-5)^2)/4-((x+3)^2)/9 =1 is (-3,7),(-3,3)