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

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

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

foci

(- 1, 2), (- 1, - 8)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (- 1, - 3), a = 4, b = 3

(- 1, - 3 + c), (- 1, - 3 - c)

Compute

c :

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

(- 1, - 3 + 5), (- 1, - 3 - 5)

Refine

(- 1, 2), (- 1, - 8)

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