What is the foci ((y+5)^2)/(1/9)-((x-3)^2)/(1/4)=1 ?

The foci ((y+5)^2)/(1/9)-((x-3)^2)/(1/4)=1 is (3,(-30+sqrt(13))/6),(3,(-30-sqrt(13))/6)

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foci ((y+5)^2)/(1/9)-((x-3)^2)/(1/4)=1

foci

\frac{( y + 5 ) ^{2}}{\frac{1}{9}} - \frac{( x - 3 ) ^{2}}{\frac{1}{4}} = 1

(3, \frac{- 30 + 13}{6}), (3, \frac{- 30 - 13}{6})

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 5 ) ^{2}}{\frac{1}{9}} - \frac{( x - 3 ) ^{2}}{\frac{1}{4}} = 1 :

Up-down Hyperbola with

(h, k) = (3, - 5), a = \frac{1}{3}, b = \frac{1}{2}

(3, - 5 + c), (3, - 5 - c)

Compute

c :

c = (\frac{1}{3})^{2} + (\frac{1}{2})^{2} : \frac{13}{6}

(3, - 5 + \frac{13}{6}), (3, - 5 - \frac{13}{6})

Refine

(3, \frac{- 30 + 13}{6}), (3, \frac{- 30 - 13}{6})

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

\frac{( x + 3 ) ^{2}}{25} - \frac{( y - 2 ) ^{2}}{49} = 1

4 y^{2} - x^{2} = 900

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

\frac{x ^{2}}{64} - \frac{y ^{2}}{32} = 1

What is the foci ((y+5)^2)/(1/9)-((x-3)^2)/(1/4)=1 ?
The foci ((y+5)^2)/(1/9)-((x-3)^2)/(1/4)=1 is (3,(-30+sqrt(13))/6),(3,(-30-sqrt(13))/6)