What is the foci ((x-3)^2)/4+((y+1)^2)/9 =1 ?

The foci ((x-3)^2)/4+((y+1)^2)/9 =1 is (3,-1+sqrt(5)),(3,-1-sqrt(5))

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foci ((x-3)^2)/4+((y+1)^2)/9 =1

foci

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1

(3, - 1 + 5), (3, - 1 - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (3, - 1), b = 3, a = 2

(3, - 1 + c), (3, - 1 - c)

Compute

c :

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

(3, - 1 + 5), (3, - 1 - 5)

\frac{x ^{2}}{15} + \frac{y ^{2}}{64} = 1

25 x^{2} + 9 y^{2} - 150 x - 18 y + 9 = 0

\frac{x ^{2}}{64} + \frac{y ^{2}}{4} = 1

\frac{( x - 1 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{25} = 1

16 x^{2} + 64 x + 4 y^{2} - 8 y + 4 = 0

What is the foci ((x-3)^2)/4+((y+1)^2)/9 =1 ?
The foci ((x-3)^2)/4+((y+1)^2)/9 =1 is (3,-1+sqrt(5)),(3,-1-sqrt(5))