What is the foci ((y+3)^2)/9-((x-2)^2)/(16)=1 ?

The foci ((y+3)^2)/9-((x-2)^2)/(16)=1 is (2,2),(2,-8)

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foci ((y+3)^2)/9-((x-2)^2)/(16)=1

foci

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 2 ) ^{2}}{16} = 1

(2, 2), (2, - 8)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 2 ) ^{2}}{16} = 1 :

Up-down Hyperbola with

(h, k) = (2, - 3), a = 3, b = 4

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

Compute

c :

c = 3^{2} + 4^{2} : 5

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

Refine

(2, 2), (2, - 8)

4 y^{2} + 48 x - 20 y = 71

y^{2} + 4 x = 0

y^{2} = - 9 x

x^{2} + y^{2} = 36

y = x^{2}

What is the foci ((y+3)^2)/9-((x-2)^2)/(16)=1 ?
The foci ((y+3)^2)/9-((x-2)^2)/(16)=1 is (2,2),(2,-8)