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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

\frac{x ^{2}}{3} + \frac{y ^{2}}{2} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{81} = 1

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{9} = 1

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

\frac{x ^{2}}{16} + \frac{y ^{2}}{36} = 1

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