What is the foci ((x+2)^2)/9+((y-3)^2)/(16)=1 ?

The foci ((x+2)^2)/9+((y-3)^2)/(16)=1 is (-2,3+sqrt(7)),(-2,3-sqrt(7))

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foci ((x+2)^2)/9+((y-3)^2)/(16)=1

foci

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

(- 2, 3 + 7), (- 2, 3 - 7)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, 3), b = 4, a = 3

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

Compute

c :

c = 4^{2} - 3^{2} : 7

(- 2, 3 + 7), (- 2, 3 - 7)

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{36} - \frac{y ^{2}}{64} = 1

x^{2} + y^{2} - 25 = 0

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1

What is the foci ((x+2)^2)/9+((y-3)^2)/(16)=1 ?
The foci ((x+2)^2)/9+((y-3)^2)/(16)=1 is (-2,3+sqrt(7)),(-2,3-sqrt(7))