What is the foci ((y+3)^2)/9-((x-1)^2)/(18)=1 ?

The foci ((y+3)^2)/9-((x-1)^2)/(18)=1 is (1,-3+3sqrt(3)),(1,-3-3sqrt(3))

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foci ((y+3)^2)/9-((x-1)^2)/(18)=1

foci

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{18} = 1

(1, - 3 + 33), (1, - 3 - 33)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{18} = 1 :

Up-down Hyperbola with

(h, k) = (1, - 3), a = 3, b = 32

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

Compute

c :

c = 3^{2} + (32)^{2} : 33

(1, - 3 + 33), (1, - 3 - 33)

\frac{( y - 3 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{27} = 1

y^{2} - 4 y - 4 x^{2} + 13 = 0

\frac{( x - 3 ) ^{2}}{9} - \frac{( y + 10 ) ^{2}}{100} = 1

4 x^{2} - 12 y^{2} - 24 x + 24 y + 51 = 0

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

What is the foci ((y+3)^2)/9-((x-1)^2)/(18)=1 ?
The foci ((y+3)^2)/9-((x-1)^2)/(18)=1 is (1,-3+3sqrt(3)),(1,-3-3sqrt(3))