What is the foci (y^2)/(16)-(x^2)/(25)=1 ?

The foci (y^2)/(16)-(x^2)/(25)=1 is (0,sqrt(41)),(0,-sqrt(41))

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foci (y^2)/(16)-(x^2)/(25)=1

foci

(0, 41), (0, - 41)

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{16} - \frac{x ^{2}}{25} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 4, b = 5

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

Compute

c :

c = 4^{2} + 5^{2} : 41

(0, 0 + 41), (0, 0 - 41)

Refine

(0, 41), (0, - 41)

What is the foci (y^2)/(16)-(x^2)/(25)=1 ?
The foci (y^2)/(16)-(x^2)/(25)=1 is (0,sqrt(41)),(0,-sqrt(41))