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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

16 x^{2} - 96 x + 9 y^{2} = 0

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

81 x^{2} - 1296 x + 16 y^{2} + 64 y = - 3952

x^{2} + 8 y^{2} = 32

25 x^{2} + y^{2} - 100 x + 8 y + 91 = 0

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