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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

7 x^{2} + 16 y^{2} = 112

\frac{x ^{2}}{5} + \frac{y ^{2}}{3} = 1

6 x^{2} + 8 y^{2} + 12 x - 32 y = 178

9 x^{2} + 49 y^{2} - 144 x + 490 y + 1360 = 0

9 x^{2} + 18 x + 4 y^{2} - 24 y + 9 = 0

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