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

The foci ((x+5)^2)/5+((y+9)^2)/9 =1 is (-5,-7),(-5,-11)

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

foci

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

(- 5, - 7), (- 5, - 11)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 5, - 9), b = 3, a = 5

(- 5, - 9 + c), (- 5, - 9 - c)

Compute

c :

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

(- 5, - 9 + 2), (- 5, - 9 - 2)

Refine

(- 5, - 7), (- 5, - 11)

\frac{( x + 7 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{9} = 1

16 x^{2} + 9 y^{2} - 96 x + 36 y - 36 = 0

\frac{x ^{2}}{100} + \frac{y ^{2}}{44} = 1

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

4 x^{2} + 8 y^{2} = 3

What is the foci ((x+5)^2)/5+((y+9)^2)/9 =1 ?
The foci ((x+5)^2)/5+((y+9)^2)/9 =1 is (-5,-7),(-5,-11)