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)

81 x^{2} + 972 x - 16 y^{2} - 128 y + 1364 = 0

36 x^{2} - 4 y^{2} = 144

(\frac{y}{49})^{2} - (\frac{x}{36})^{2} = 1

25 y^{2} - 64 x^{2} = 1600

4 y^{2} - 16 x^{2} = 64

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)