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

The foci ((x+2)^2)/(2^2)+((y-3)^2)/(3^2)=1 is (-2,3+sqrt(5)),(-2,3-sqrt(5))

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

foci

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

(- 2, 3 + 5), (- 2, 3 - 5)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

(- 2, 3 + 5), (- 2, 3 - 5)

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

25 x^{2} + 4 y^{2} + 100 x = 0

\frac{x ^{2}}{9 ^{2}} + \frac{y ^{2}}{6 ^{2}} = 1

\frac{x ^{2}}{5 ^{2}} + \frac{y ^{2}}{1 3 ^{2}} = 1

\frac{x ^{2}}{1} + \frac{y ^{2}}{5} = 1

What is the foci ((x+2)^2)/(2^2)+((y-3)^2)/(3^2)=1 ?
The foci ((x+2)^2)/(2^2)+((y-3)^2)/(3^2)=1 is (-2,3+sqrt(5)),(-2,3-sqrt(5))