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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

\frac{x ^{2}}{25} + \frac{y ^{2}}{169} = 1

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

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

81 x^{2} - 1296 x + 16 y^{2} + 64 y = - 3952

x^{2} + 8 y^{2} = 32

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