What is the foci ((x-3)^2)/9+((y+2)^2)/(25)=1 ?

The foci ((x-3)^2)/9+((y+2)^2)/(25)=1 is (3,2),(3,-6)

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foci ((x-3)^2)/9+((y+2)^2)/(25)=1

foci

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1

(3, 2), (3, - 6)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1 :

Ellipse with center

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

(3, - 2 + c), (3, - 2 - c)

Compute

c :

c = 5^{2} - 3^{2} : 4

(3, - 2 + 4), (3, - 2 - 4)

Refine

(3, 2), (3, - 6)

25 \frac{x ^{2}}{4} - 50 x + 100 y^{2} + 800 y + 1075 = 0

x^{2} + 16 y^{2} = 1

28 x^{2} + 64 y^{2} - 56 x - 420 = 0

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{10} = 1

\frac{x ^{2}}{16} + \frac{( y - 1 ) ^{2}}{49} = 1

What is the foci ((x-3)^2)/9+((y+2)^2)/(25)=1 ?
The foci ((x-3)^2)/9+((y+2)^2)/(25)=1 is (3,2),(3,-6)