What is the foci ((x-2)^2)/(8^2)+((y+2)^2)/(3^2)=1 ?

The foci ((x-2)^2)/(8^2)+((y+2)^2)/(3^2)=1 is (2+sqrt(55),-2),(2-sqrt(55),-2)

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foci ((x-2)^2)/(8^2)+((y+2)^2)/(3^2)=1

foci

\frac{( x - 2 ) ^{2}}{8 ^{2}} + \frac{( y + 2 ) ^{2}}{3 ^{2}} = 1

(2 + 55, - 2), (2 - 55, - 2)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{8 ^{2}} + \frac{( y + 2 ) ^{2}}{3 ^{2}} = 1 :

Ellipse with center

(h, k) = (2, - 2), a = 8, b = 3

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

Compute

c :

c = 8^{2} - 3^{2} : 55

(2 + 55, - 2), (2 - 55, - 2)

y^{2} = 1 - 49 x^{2}

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

9 x^{2} + 49 y^{2} + 18 x + 294 y - 9 = 0

9 x^{2} + 7 y^{2} - 63 = 0

16 x^{2} + 64 y^{2} = 128

What is the foci ((x-2)^2)/(8^2)+((y+2)^2)/(3^2)=1 ?
The foci ((x-2)^2)/(8^2)+((y+2)^2)/(3^2)=1 is (2+sqrt(55),-2),(2-sqrt(55),-2)