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

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

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foci ((x-2)^2)/9+((y+5)^2)/4 =1

foci

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 5 ) ^{2}}{4} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 5 ) ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (2, - 5), a = 3, b = 2

(2 + c, - 5), (2 - c, - 5)

Compute

c :

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

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

\frac{x ^{2}}{5 ^{2}} + \frac{y ^{2}}{4 ^{2}} = 1

\frac{( x - 1 ) ^{2}}{81} + \frac{( y - 7 ) ^{2}}{45} = 1

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

x^{2} + y^{2} + 10 x - 6 y + 33 = 0

\frac{( x - 2 ) ^{2}}{36} + \frac{( y - 6 ) ^{2}}{4} = 1

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