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

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

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

foci

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Compute

c :

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

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

Refine

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

\frac{x ^{2}}{36} + \frac{y ^{2}}{25} = 1

9 x^{2} - 16 y^{2} = 144

x^{2} + y^{2} + 6 x - 4 y = 0

f (x) = x^{2} + 2 x + 1

y^{2} = 28 x

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