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

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

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

foci

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

(- 2, 1 + 119), (- 2, 1 - 119)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, 1), b = 12, a = 5

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

Compute

c :

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

(- 2, 1 + 119), (- 2, 1 - 119)

\frac{( x - 4 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1

25 x^{2} + 9 y^{2} - 100 x + 18 y - 116 = 0

\frac{x ^{2}}{81} + \frac{y ^{2}}{49} = 1

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1

x^{2} + 2 y^{2} - 2 x + 8 y + 5 = 0

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