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

The foci ((y-2)^2)/9-((x-1)^2)/(16)=1 is (1,7),(1,-3)

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

foci

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

(1, 7), (1, - 3)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (1, 2), a = 3, b = 4

(1, 2 + c), (1, 2 - c)

Compute

c :

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

(1, 2 + 5), (1, 2 - 5)

Refine

(1, 7), (1, - 3)

x^{2} - 3 y^{2} + 12 = 0

9 x^{2} - 9 y^{2} - 18 x + 54 = 0

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

\frac{y ^{2}}{36} - \frac{x ^{2}}{81} = 1

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

What is the foci ((y-2)^2)/9-((x-1)^2)/(16)=1 ?
The foci ((y-2)^2)/9-((x-1)^2)/(16)=1 is (1,7),(1,-3)