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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

\frac{( x + 5 ) ^{2}}{20} + \frac{( y - 6 ) ^{2}}{11} = 1

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

\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

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