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

The foci ((x-2)^2)/2+((y-1)^2)/8 =1 is (2,1+sqrt(6)),(2,1-sqrt(6))

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

foci

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

(2, 1 + 6), (2, 1 - 6)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, 1), b = 22, a = 2

(2, 1 + c), (2, 1 - c)

Compute

c :

c = (22)^{2} - 2^{2} : 6

(2, 1 + 6), (2, 1 - 6)

9 x^{2} + 4 y^{2} = 25

4 x^{2} + 9 y^{2} + 24 x - 36 y + 36 = 0

4 x^{2} + 16 y^{2} - 16 x - 48 = 0

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

4 x^{2} + 9 y^{2} - 48 x + 72 y = - 144

What is the foci ((x-2)^2)/2+((y-1)^2)/8 =1 ?
The foci ((x-2)^2)/2+((y-1)^2)/8 =1 is (2,1+sqrt(6)),(2,1-sqrt(6))