What is the foci ((x+7)^2}{16}+\frac{(y+3)^2)/9 =1 ?

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 7, - 3), a = 4, b = 3

(- 7 + c, - 3), (- 7 - c, - 3)

Compute

c :

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

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

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

16 x^{2} + 4 y^{2} - 64 = 0

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