What is the foci ((y-1)^2)/9-((x-3)^2)/(16)=1 ?

The foci ((y-1)^2)/9-((x-3)^2)/(16)=1 is (3,6),(3,-4)

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foci ((y-1)^2)/9-((x-3)^2)/(16)=1

foci

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

(3, 6), (3, - 4)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Compute

c :

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

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

Refine

(3, 6), (3, - 4)

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

\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

What is the foci ((y-1)^2)/9-((x-3)^2)/(16)=1 ?
The foci ((y-1)^2)/9-((x-3)^2)/(16)=1 is (3,6),(3,-4)