What is the foci ((x+1)^2}{10}+\frac{(y-3)^2)/9 =1 ?

The foci ((x+1)^2}{10}+\frac{(y-3)^2)/9 =1 is (0,3),(-2,3)

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foci ((x+1)^2}{10}+\frac{(y-3)^2)/9 =1

foci

\frac{( x + 1 ) ^{2}}{10} + \frac{( y - 3 ) ^{2}}{9} = 1

(0, 3), (- 2, 3)

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{10} + \frac{( y - 3 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (- 1, 3), a = 10, b = 3

(- 1 + c, 3), (- 1 - c, 3)

Compute

c :

c = 10^{2} - 3^{2} : 1

(- 1 + 1, 3), (- 1 - 1, 3)

Refine

(0, 3), (- 2, 3)

\frac{( x ) ^{2}}{44} + \frac{y ^{2}}{125} = 1

2 x^{2} + 8 y^{2} = 16

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

25 x^{2} + 9 y^{2} - 100 x - 54 y - 44 = 0

25 x^{2} + 16 y^{2} + 100 x - 300 = 0

What is the foci ((x+1)^2}{10}+\frac{(y-3)^2)/9 =1 ?
The foci ((x+1)^2}{10}+\frac{(y-3)^2)/9 =1 is (0,3),(-2,3)