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

The foci (y^2)/9-(x^2)/(16)=1 is (0,5),(0,-5)

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

foci

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

(0, 5), (0, - 5)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (0, 0), a = 3, b = 4

(0, 0 + c), (0, 0 - c)

Compute

c :

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

(0, 0 + 5), (0, 0 - 5)

Refine

(0, 5), (0, - 5)

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = 1

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

x^{2} = - 6 y

y^{2} - 4 y - 4 x = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1

What is the foci (y^2)/9-(x^2)/(16)=1 ?
The foci (y^2)/9-(x^2)/(16)=1 is (0,5),(0,-5)