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

The foci ((x+2)^2)/(25)+((y-1)^2)/(169)=1 is (-2,13),(-2,-11)

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

foci

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

(- 2, 13), (- 2, - 11)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, 1), b = 13, a = 5

(- 2, 1 + c), (- 2, 1 - c)

Compute

c :

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

(- 2, 1 + 12), (- 2, 1 - 12)

Refine

(- 2, 13), (- 2, - 11)

\frac{( x - 3 ) ^{2}}{25} + \frac{y ^{2}}{9} = 1

x^{2} + 4 y^{2} = 16

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1

25 x^{2} - 200 x + y^{2} + 4 y + 379 = 0

64 x^{2} + 100 y^{2} + 768 x - 600 y - 3196 = 0

What is the foci ((x+2)^2)/(25)+((y-1)^2)/(169)=1 ?
The foci ((x+2)^2)/(25)+((y-1)^2)/(169)=1 is (-2,13),(-2,-11)