What is the vertices ((y-2)^2)/9-((x+1)^2)/(25)=1 ?

The vertices ((y-2)^2)/9-((x+1)^2)/(25)=1 is (-1,5),(-1,-1)

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vertices ((y-2)^2)/9-((x+1)^2)/(25)=1

vertices

\frac{( y - 2 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{25} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 2 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{25} = 1 :

Up-down Hyperbola with

(h, k) = (- 1, 2), a = 3, b = 5

(- 1, 2 + 3), (- 1, 2 - 3)

Refine

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

9 y^{2} - x^{2} + 54 y + 2 x + 62 = 0

x^{2} - y^{2} = 6

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

\frac{( x - 3 ) ^{2}}{3} - \frac{( y - 2 ) ^{2}}{22} = 1

16 x^{2} - 9 y^{2} + 64 x - 90 y = 305

What is the vertices ((y-2)^2)/9-((x+1)^2)/(25)=1 ?
The vertices ((y-2)^2)/9-((x+1)^2)/(25)=1 is (-1,5),(-1,-1)