What is the foci ((x-5)^2)/9+((y+2)^2)/(25)=1 ?

The foci ((x-5)^2)/9+((y+2)^2)/(25)=1 is (5,2),(5,-6)

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foci ((x-5)^2)/9+((y+2)^2)/(25)=1

foci

\frac{( x - 5 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1

(5, 2), (5, - 6)

Solution steps

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

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1 :

Ellipse with center

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

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

Compute

c :

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

(5, - 2 + 4), (5, - 2 - 4)

Refine

(5, 2), (5, - 6)

\frac{x ^{2}}{0.38 3 ^{2}} + \frac{y ^{2}}{0.374 8 ^{2}} = 1

\frac{x ^{2}}{140} + \frac{( y - 6 ) ^{2}}{4} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{4} = 1

25 x^{2} + 16 y^{2} - 150 x - 32 y - 159 = 0

\frac{( x + 6 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{25} = 1

What is the foci ((x-5)^2)/9+((y+2)^2)/(25)=1 ?
The foci ((x-5)^2)/9+((y+2)^2)/(25)=1 is (5,2),(5,-6)