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

The foci ((x+4)^2)/5+((y-5)^2)/9 =1 is (-4,7),(-4,3)

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

foci

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

(- 4, 7), (- 4, 3)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

(- 4, 7), (- 4, 3)

\frac{x ^{2}}{1} + \frac{y ^{2}}{64} = 1

\frac{( x ) ^{2}}{13} + \frac{y ^{2}}{4} = 1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{16} = 1

\frac{1 x ^{2}}{16} + \frac{y ^{2}}{9} = 1

\frac{( x - 4 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1

What is the foci ((x+4)^2)/5+((y-5)^2)/9 =1 ?
The foci ((x+4)^2)/5+((y-5)^2)/9 =1 is (-4,7),(-4,3)