What is the foci ((y+2)^2}{25}-\frac{(x-3)^2)/4 =1 ?

The foci ((y+2)^2}{25}-\frac{(x-3)^2)/4 =1 is (3,-2+sqrt(29)),(3,-2-sqrt(29))

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foci ((y+2)^2}{25}-\frac{(x-3)^2)/4 =1

foci

\frac{( y + 2 ) ^{2}}{25} - \frac{( x - 3 ) ^{2}}{4} = 1

(3, - 2 + 29), (3, - 2 - 29)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 2 ) ^{2}}{25} - \frac{( x - 3 ) ^{2}}{4} = 1 :

Up-down Hyperbola with

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

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

Compute

c :

c = 5^{2} + 2^{2} : 29

(3, - 2 + 29), (3, - 2 - 29)

25 x^{2} - 9 y^{2} = 225

\frac{( y - 5 ) ^{2}}{16} - \frac{( x - 4 ) ^{2}}{9} = 1

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

144 y^{2} - 25 x^{2} = 3600

\frac{y ^{2}}{25} - \frac{x ^{2}}{39} = 1

What is the foci ((y+2)^2}{25}-\frac{(x-3)^2)/4 =1 ?
The foci ((y+2)^2}{25}-\frac{(x-3)^2)/4 =1 is (3,-2+sqrt(29)),(3,-2-sqrt(29))