What is the foci ((x-8)^2)/(64)+((y-5)^2)/(5^2)=1 ?

The foci ((x-8)^2)/(64)+((y-5)^2)/(5^2)=1 is (8+sqrt(39),5),(8-sqrt(39),5)

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foci ((x-8)^2)/(64)+((y-5)^2)/(5^2)=1

foci

\frac{( x - 8 ) ^{2}}{64} + \frac{( y - 5 ) ^{2}}{5 ^{2}} = 1

(8 + 39, 5), (8 - 39, 5)

Solution steps

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

Calculate ellipse properties

\frac{( x - 8 ) ^{2}}{64} + \frac{( y - 5 ) ^{2}}{5 ^{2}} = 1 :

Ellipse with center

(h, k) = (8, 5), a = 8, b = 5

(8 + c, 5), (8 - c, 5)

Compute

c :

c = 8^{2} - 5^{2} : 39

(8 + 39, 5), (8 - 39, 5)

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0

25 x^{2} + 64 y^{2} = 1600

\frac{( x + 2 ) ^{2}}{22} + \frac{( y + 3 ) ^{2}}{18} = 1

25 (x + 2)^{2} + 4 (y - 3)^{2} = 100

\frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

What is the foci ((x-8)^2)/(64)+((y-5)^2)/(5^2)=1 ?
The foci ((x-8)^2)/(64)+((y-5)^2)/(5^2)=1 is (8+sqrt(39),5),(8-sqrt(39),5)