What is the foci ((x-8)^2)/2+((y-3)^2)/(27)=1 ?

The foci ((x-8)^2)/2+((y-3)^2)/(27)=1 is (8,8),(8,-2)

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foci ((x-8)^2)/2+((y-3)^2)/(27)=1

foci

\frac{( x - 8 ) ^{2}}{2} + \frac{( y - 3 ) ^{2}}{27} = 1

(8, 8), (8, - 2)

Solution steps

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

Calculate ellipse properties

\frac{( x - 8 ) ^{2}}{2} + \frac{( y - 3 ) ^{2}}{27} = 1 :

Ellipse with center

(h, k) = (8, 3), b = 33, a = 2

(8, 3 + c), (8, 3 - c)

Compute

c :

c = (33)^{2} - 2^{2} : 5

(8, 3 + 5), (8, 3 - 5)

Refine

(8, 8), (8, - 2)

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

3 x^{2} + y^{2} = 6

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

\frac{( x ^{2} )}{49} + \frac{( y ^{2} )}{25} = 1

9 x^{2} + 4 y^{2} + 32 y + 28 = 0

What is the foci ((x-8)^2)/2+((y-3)^2)/(27)=1 ?
The foci ((x-8)^2)/2+((y-3)^2)/(27)=1 is (8,8),(8,-2)