What is the foci (y^2)/(49)-(x^2)/(16)=1 ?

The foci (y^2)/(49)-(x^2)/(16)=1 is (0,sqrt(65)),(0,-sqrt(65))

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foci (y^2)/(49)-(x^2)/(16)=1

foci

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

(0, 65), (0, - 65)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (0, 0), a = 7, b = 4

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

Compute

c :

c = 7^{2} + 4^{2} : 65

(0, 0 + 65), (0, 0 - 65)

Refine

(0, 65), (0, - 65)

25 x^{2} - 16 y^{2} + 250 x + 32 y + 109 = 0

\frac{( x + 2 ) ^{2}}{25} - \frac{( y - 1 ) ^{2}}{169} = 1

\frac{( x - 2 ) ^{2}}{5} - \frac{( y + 1 ) ^{2}}{9} = 1

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

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

What is the foci (y^2)/(49)-(x^2)/(16)=1 ?
The foci (y^2)/(49)-(x^2)/(16)=1 is (0,sqrt(65)),(0,-sqrt(65))