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

The foci ((x-3)^2)/4+((y+2)^2)/9 =3 is (3,-2+sqrt(15)),(3,-2-sqrt(15))

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

foci

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

(3, - 2 + 15), (3, - 2 - 15)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (3, - 2), b = 33, a = 23

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

Compute

c :

c = (33)^{2} - (23)^{2} : 15

(3, - 2 + 15), (3, - 2 - 15)

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

x^{2} + 4 y^{2} - 4 x - 8 y - 92 = 0

x^{2} + 9 y^{2} = 36

9 x^{2} + 25 y^{2} + 108 x - 200 y + 499 = 0

\frac{( x - 6 ) ^{2}}{49} + \frac{( y - 6 ) ^{2}}{36} = 1

What is the foci ((x-3)^2)/4+((y+2)^2)/9 =3 ?
The foci ((x-3)^2)/4+((y+2)^2)/9 =3 is (3,-2+sqrt(15)),(3,-2-sqrt(15))