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

The foci ((y-2)^2)/4-((x+3)^2)/9 =1 is (-3,2+sqrt(13)),(-3,2-sqrt(13))

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

foci

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

(- 3, 2 + 13), (- 3, 2 - 13)

Solution steps

(h, k + c), (h, k - c)

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

(- 3, 2 + c), (- 3, 2 - c)

Compute

c :

c = 2^{2} + 3^{2} : 13

(- 3, 2 + 13), (- 3, 2 - 13)

16 x^{2} - 20 y^{2} - 320 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{121} = 1

5 y^{2} - 4 x^{2} - 80 = 0

y^{2} + 2 y = 4 x^{2} + 3

9 x^{2} - 4 y^{2} - 18 x + 16 y - 43 = 0

What is the foci ((y-2)^2)/4-((x+3)^2)/9 =1 ?
The foci ((y-2)^2)/4-((x+3)^2)/9 =1 is (-3,2+sqrt(13)),(-3,2-sqrt(13))