What is the foci ((x-5)^2)/(8^2)+((y+2)^2)/(10^2)=1 ?

The foci ((x-5)^2)/(8^2)+((y+2)^2)/(10^2)=1 is (5,4),(5,-8)

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foci ((x-5)^2)/(8^2)+((y+2)^2)/(10^2)=1

foci

\frac{( x - 5 ) ^{2}}{8 ^{2}} + \frac{( y + 2 ) ^{2}}{1 0 ^{2}} = 1

(5, 4), (5, - 8)

Solution steps

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

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{8 ^{2}} + \frac{( y + 2 ) ^{2}}{1 0 ^{2}} = 1 :

Ellipse with center

(h, k) = (5, - 2), b = 10, a = 8

(5, - 2 + c), (5, - 2 - c)

Compute

c :

c = 1 0^{2} - 8^{2} : 6

(5, - 2 + 6), (5, - 2 - 6)

Refine

(5, 4), (5, - 8)

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

9 x^{2} - 54 x + 4 y^{2} = - 45

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{144} = 1

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

25 x^{2} + 9 y^{2} - 100 x + 18 y - 116 = 0

What is the foci ((x-5)^2)/(8^2)+((y+2)^2)/(10^2)=1 ?
The foci ((x-5)^2)/(8^2)+((y+2)^2)/(10^2)=1 is (5,4),(5,-8)