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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (4, 5), a = 3, b = 2

(4 + c, 5), (4 - c, 5)

Compute

c :

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

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

4 x^{2} + 9 y^{2} + 12 x - 108 y + 297 = 0

9 x^{(2)} + 16 y^{(2)} - 54 x + 64 y + 1 = 0

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

2 x^{2} + 4 y^{2} + 8 x + 9 y - 3 = 0

36 x^{2} + 4 y^{2} + 360 x - 32 y + 820 = 0

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