What is the foci (y^2)/(12^2)-(x^2)/(5^2)=1 ?

The foci (y^2)/(12^2)-(x^2)/(5^2)=1 is (0,13),(0,-13)

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foci (y^2)/(12^2)-(x^2)/(5^2)=1

foci

\frac{y ^{2}}{1 2 ^{2}} - \frac{x ^{2}}{5 ^{2}} = 1

(0, 13), (0, - 13)

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{1 2 ^{2}} - \frac{x ^{2}}{5 ^{2}} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 12, b = 5

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

Compute

c :

c = 1 2^{2} + 5^{2} : 13

(0, 0 + 13), (0, 0 - 13)

Refine

(0, 13), (0, - 13)

y^{2} - 4 x^{2} - 10 y - 40 x - 100 = 0

- 1 = x^{2} - y^{2}

\frac{( x ^{2} )}{4} - 0.15 y^{2} = 1

9 x^{2} - y^{2} - 36 x - 6 y + 18 = 0

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

What is the foci (y^2)/(12^2)-(x^2)/(5^2)=1 ?
The foci (y^2)/(12^2)-(x^2)/(5^2)=1 is (0,13),(0,-13)