What is the foci ((x-4)^2)/9+((y+6)^2)/4 =1 ?

The foci ((x-4)^2)/9+((y+6)^2)/4 =1 is (4+sqrt(5),-6),(4-sqrt(5),-6)

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foci ((x-4)^2)/9+((y+6)^2)/4 =1

foci

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{4} = 1

(4 + 5, - 6), (4 - 5, - 6)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (4, - 6), a = 3, b = 2

(4 + c, - 6), (4 - c, - 6)

Compute

c :

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

(4 + 5, - 6), (4 - 5, - 6)

\frac{x ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

\frac{( x + 4 ) ^{2}}{225} + \frac{( y - 7 ) ^{2}}{81} = 1

9 x^{2} + 4 y^{2} = 24 y

25 x^{2} + 16 y^{2} + 100 x - 96 y - 156 = 0

2 x^{2} + 4 y^{2} - 8 x - 4 y - 15 = 0

What is the foci ((x-4)^2)/9+((y+6)^2)/4 =1 ?
The foci ((x-4)^2)/9+((y+6)^2)/4 =1 is (4+sqrt(5),-6),(4-sqrt(5),-6)