What is the foci ((x+2)^2}{10}+\frac{(y-5)^2)/2 =1 ?

The foci ((x+2)^2}{10}+\frac{(y-5)^2)/2 =1 is (-2+2sqrt(2),5),(-2-2sqrt(2),5)

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foci ((x+2)^2}{10}+\frac{(y-5)^2)/2 =1

foci

\frac{( x + 2 ) ^{2}}{10} + \frac{( y - 5 ) ^{2}}{2} = 1

(- 2 + 22, 5), (- 2 - 22, 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 2 ) ^{2}}{10} + \frac{( y - 5 ) ^{2}}{2} = 1 :

Ellipse with center

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

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

Compute

c :

c = 10^{2} - 2^{2} : 22

(- 2 + 22, 5), (- 2 - 22, 5)

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{81} = 1

4 x^{2} + 25 y^{2} - 100 = 0

25 x^{2} + 9 y^{2} - 225 = 0

\frac{( x - 1 ) ^{2}}{36} + \frac{( y - 1 ) ^{2}}{25} = 1

9 x^{2} + 16 y^{2} + 72 x + 128 y + 256 = 0

What is the foci ((x+2)^2}{10}+\frac{(y-5)^2)/2 =1 ?
The foci ((x+2)^2}{10}+\frac{(y-5)^2)/2 =1 is (-2+2sqrt(2),5),(-2-2sqrt(2),5)