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

The foci ((x+6)^2)/9+((y-3)^2)/(25)=1 is (-6,7),(-6,-1)

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

foci

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

(- 6, 7), (- 6, - 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 6, 3), b = 5, a = 3

(- 6, 3 + c), (- 6, 3 - c)

Compute

c :

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

(- 6, 3 + 4), (- 6, 3 - 4)

Refine

(- 6, 7), (- 6, - 1)

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

\frac{x ^{2}}{32} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{100} + \frac{( y - 6 ) ^{2}}{64} = 1

6 x^{2} + 9 y^{2} - 18 y - 45 = 0

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

What is the foci ((x+6)^2)/9+((y-3)^2)/(25)=1 ?
The foci ((x+6)^2)/9+((y-3)^2)/(25)=1 is (-6,7),(-6,-1)