What is the foci ((y-2)^2}{40}-\frac{(x-4)^2)/9 =1 ?

The foci ((y-2)^2}{40}-\frac{(x-4)^2)/9 =1 is (4,9),(4,-5)

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foci ((y-2)^2}{40}-\frac{(x-4)^2)/9 =1

foci

\frac{( y - 2 ) ^{2}}{40} - \frac{( x - 4 ) ^{2}}{9} = 1

(4, 9), (4, - 5)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 2 ) ^{2}}{40} - \frac{( x - 4 ) ^{2}}{9} = 1 :

Up-down Hyperbola with

(h, k) = (4, 2), a = 210, b = 3

(4, 2 + c), (4, 2 - c)

Compute

c :

c = (210)^{2} + 3^{2} : 7

(4, 2 + 7), (4, 2 - 7)

Refine

(4, 9), (4, - 5)

5 y^{2} - 16 x^{2} + 400 = 0

\frac{x ^{2}}{36} - \frac{( y - 10 ) ^{2}}{64} = 1

\frac{( y - 1 ) ^{2}}{49} - \frac{( x - 2 ) ^{2}}{32} = 1

\frac{( x - 3 ) ^{2}}{13} - \frac{( y - 5 ) ^{2}}{7} = 1

\frac{y ^{2}}{10} - \frac{( x - 1 ) ^{2}}{10} = 1

What is the foci ((y-2)^2}{40}-\frac{(x-4)^2)/9 =1 ?
The foci ((y-2)^2}{40}-\frac{(x-4)^2)/9 =1 is (4,9),(4,-5)