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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

\frac{x ^{2}}{16} - \frac{y ^{2}}{36} = 1

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

y^{2} - 6 y - 4 x + 5 = 0

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

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

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