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

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

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

foci

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

(2 + 5, - 1), (2 - 5, - 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(2 + c, - 1), (2 - c, - 1)

Compute

c :

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

(2 + 5, - 1), (2 - 5, - 1)

7 x^{2} = 21 - 3 y^{2}

9 x^{2} + 25 y^{2} + 18 x + 50 y - 191 = 0

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

3 x^{2} - 6 x + y^{2} + 4 y - 11 = 0

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

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