What is the foci (y^2)/4-(x^2)/(16)=1 ?

The foci (y^2)/4-(x^2)/(16)=1 is (0,2sqrt(5)),(0,-2sqrt(5))

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foci (y^2)/4-(x^2)/(16)=1

foci

\frac{y ^{2}}{4} - \frac{x ^{2}}{16} = 1

(0, 25), (0, - 25)

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{4} - \frac{x ^{2}}{16} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 2, b = 4

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

Compute

c :

c = 2^{2} + 4^{2} : 25

(0, 0 + 25), (0, 0 - 25)

Refine

(0, 25), (0, - 25)

\frac{( y + 3 ) ^{2}}{64} - \frac{( x + 6 ) ^{2}}{49} = 1

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

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

x^{2} - 6 x - 6 = y^{2} + 6 y + 6

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

What is the foci (y^2)/4-(x^2)/(16)=1 ?
The foci (y^2)/4-(x^2)/(16)=1 is (0,2sqrt(5)),(0,-2sqrt(5))