What is the foci ((y-5)^2}{16}-\frac{(x-4)^2)/9 =1 ?

The foci ((y-5)^2}{16}-\frac{(x-4)^2)/9 =1 is (4,10),(4,0)

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foci ((y-5)^2}{16}-\frac{(x-4)^2)/9 =1

foci

\frac{( y - 5 ) ^{2}}{16} - \frac{( x - 4 ) ^{2}}{9} = 1

(4, 10), (4, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 5 ) ^{2}}{16} - \frac{( x - 4 ) ^{2}}{9} = 1 :

Up-down Hyperbola with

(h, k) = (4, 5), a = 4, b = 3

(4, 5 + c), (4, 5 - c)

Compute

c :

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

(4, 5 + 5), (4, 5 - 5)

Refine

(4, 10), (4, 0)

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

144 y^{2} - 25 x^{2} = 3600

\frac{y ^{2}}{25} - \frac{x ^{2}}{39} = 1

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

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

What is the foci ((y-5)^2}{16}-\frac{(x-4)^2)/9 =1 ?
The foci ((y-5)^2}{16}-\frac{(x-4)^2)/9 =1 is (4,10),(4,0)