What is the foci ((y+2)^2}{25}-\frac{(x+4)^2)/9 =1 ?

The foci ((y+2)^2}{25}-\frac{(x+4)^2)/9 =1 is (-4,-2+sqrt(34)),(-4,-2-sqrt(34))

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foci ((y+2)^2}{25}-\frac{(x+4)^2)/9 =1

foci

\frac{( y + 2 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{9} = 1

(- 4, - 2 + 34), (- 4, - 2 - 34)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 2 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{9} = 1 :

Up-down Hyperbola with

(h, k) = (- 4, - 2), a = 5, b = 3

(- 4, - 2 + c), (- 4, - 2 - c)

Compute

c :

c = 5^{2} + 3^{2} : 34

(- 4, - 2 + 34), (- 4, - 2 - 34)

(\frac{x - 6}{4})^{2} - (\frac{y}{4})^{2} = 1

25 y^{2} - 4 x^{2} = 100

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

4 x^{2} - 9 y^{2} + 3 x + 18 y + 91 = 0

4 x^{2} - 25 y^{2} = 100

What is the foci ((y+2)^2}{25}-\frac{(x+4)^2)/9 =1 ?
The foci ((y+2)^2}{25}-\frac{(x+4)^2)/9 =1 is (-4,-2+sqrt(34)),(-4,-2-sqrt(34))