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)

8 x^{2} + 2 y^{2} + 16 x = 0

\frac{x ^{2}}{169} + \frac{y ^{2}}{144} = 1

16 x^{2} + 25 y^{2} - 64 x - 50 y - 311 = 0

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

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