What is the foci ((y+3)^2)/(169)+((x+2)^2)/(25)=1 ?

The foci ((y+3)^2)/(169)+((x+2)^2)/(25)=1 is (-2,9),(-2,-15)

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foci ((y+3)^2)/(169)+((x+2)^2)/(25)=1

foci

\frac{( y + 3 ) ^{2}}{169} + \frac{( x + 2 ) ^{2}}{25} = 1

(- 2, 9), (- 2, - 15)

Solution steps

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

Calculate ellipse properties

\frac{( y + 3 ) ^{2}}{169} + \frac{( x + 2 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (- 2, - 3), b = 13, a = 5

(- 2, - 3 + c), (- 2, - 3 - c)

Compute

c :

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

(- 2, - 3 + 12), (- 2, - 3 - 12)

Refine

(- 2, 9), (- 2, - 15)

\frac{x ^{2}}{15} + \frac{y ^{2}}{45} = 1

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

9 x^{2} + 4 y^{2} - 18 x + 16 y - 11 = 0

\frac{x ^{2}}{99} + \frac{y ^{2}}{225} = 1

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

What is the foci ((y+3)^2)/(169)+((x+2)^2)/(25)=1 ?
The foci ((y+3)^2)/(169)+((x+2)^2)/(25)=1 is (-2,9),(-2,-15)