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

The foci ((y-2)^2)/9-((x+2)^2)/(16)=1 is (-2,7),(-2,-3)

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

foci

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

(- 2, 7), (- 2, - 3)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Compute

c :

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

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

Refine

(- 2, 7), (- 2, - 3)

12 y^{2} - 4 x^{2} + 16 x + 72 y + 44 = 0

\frac{y ^{2}}{45} - \frac{x ^{2}}{36} = 1

y^{2} - 4 x^{2} + 16 x - 2 y = 31

\frac{y ^{2}}{12.00 6 ^{2}} - \frac{x ^{2}}{2.00 1 ^{2}} = 1

\frac{x ^{2}}{81} - \frac{y ^{2}}{49} = 1

What is the foci ((y-2)^2)/9-((x+2)^2)/(16)=1 ?
The foci ((y-2)^2)/9-((x+2)^2)/(16)=1 is (-2,7),(-2,-3)