What is the foci (y^2)/(169)+(x^2)/(25)=1 ?

The foci (y^2)/(169)+(x^2)/(25)=1 is (0,12),(0,-12)

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foci (y^2)/(169)+(x^2)/(25)=1

foci

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

(0, 12), (0, - 12)

Solution steps

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

Calculate ellipse properties

\frac{y ^{2}}{169} + \frac{x ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 13, a = 5

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

Compute

c :

c = 1 3^{2} - 5^{2} : 12

(0, 0 + 12), (0, 0 - 12)

Refine

(0, 12), (0, - 12)

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

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

\frac{x ^{2}}{\frac{16}{9}} + \frac{y ^{2}}{16} = 1

(\frac{x}{5})^{2} + (\frac{y}{3})^{2} = 1

9 x^{2} + 16 y^{2} = 1

What is the foci (y^2)/(169)+(x^2)/(25)=1 ?
The foci (y^2)/(169)+(x^2)/(25)=1 is (0,12),(0,-12)