What is the foci ((y+3)^2)/(144)+((x+2)^2)/(25)=1 ?

The foci ((y+3)^2)/(144)+((x+2)^2)/(25)=1 is (-2,-3+sqrt(119)),(-2,-3-sqrt(119))

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foci ((y+3)^2)/(144)+((x+2)^2)/(25)=1

foci

\frac{( y + 3 ) ^{2}}{144} + \frac{( x + 2 ) ^{2}}{25} = 1

(- 2, - 3 + 119), (- 2, - 3 - 119)

Solution steps

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

Calculate ellipse properties

\frac{( y + 3 ) ^{2}}{144} + \frac{( x + 2 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (- 2, - 3), b = 12, a = 5

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

Compute

c :

c = 1 2^{2} - 5^{2} : 119

(- 2, - 3 + 119), (- 2, - 3 - 119)

5 x^{2} + 9 y^{2} - 20 x + 54 y + 56 = 0

25 x^{2} + 4 y^{2} - 250 x - 16 y + 541 = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{21} = 1

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

81 x^{2} + 49 y^{2} = 3969

What is the foci ((y+3)^2)/(144)+((x+2)^2)/(25)=1 ?
The foci ((y+3)^2)/(144)+((x+2)^2)/(25)=1 is (-2,-3+sqrt(119)),(-2,-3-sqrt(119))