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

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

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

foci

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Compute

c :

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

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

Refine

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

\frac{x ^{2}}{100} - \frac{y ^{2}}{49} = 1

\frac{( y - ( - 2 ) ) ^{2}}{9} - \frac{( x - 5 ) ^{2}}{27} = 1

\frac{( y - 0 ) ^{2}}{1 ^{2}} - \frac{( 3 x - 0 ) ^{2}}{90} = 1

- \frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{16} = 1

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