What is the foci 9(x+2)^2+4(y-3)^2=36 ?

The foci 9(x+2)^2+4(y-3)^2=36 is (-2,3+sqrt(5)),(-2,3-sqrt(5))

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foci 9(x+2)^2+4(y-3)^2=36

foci

9 (x + 2)^{2} + 4 (y - 3)^{2} = 36

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

Solution steps

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

Calculate ellipse properties

9 (x + 2)^{2} + 4 (y - 3)^{2} = 36 :

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)

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{169} = 1

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

\frac{x ^{2}}{121} + \frac{( y + 4 ) ^{2}}{1} = 1

9 x^{2} + 5 y^{2} = 45

16 x^{2} + 25 y^{2} + 64 x - 336 = 0

What is the foci 9(x+2)^2+4(y-3)^2=36 ?
The foci 9(x+2)^2+4(y-3)^2=36 is (-2,3+sqrt(5)),(-2,3-sqrt(5))