What is the foci ((x+2)^2)/(22)+((y+3)^2)/(18)=1 ?

The foci ((x+2)^2)/(22)+((y+3)^2)/(18)=1 is (0,-3),(-4,-3)

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foci ((x+2)^2)/(22)+((y+3)^2)/(18)=1

foci

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

(0, - 3), (- 4, - 3)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, - 3), a = 22, b = 32

(- 2 + c, - 3), (- 2 - c, - 3)

Compute

c :

c = 22^{2} - (32)^{2} : 2

(- 2 + 2, - 3), (- 2 - 2, - 3)

Refine

(0, - 3), (- 4, - 3)

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

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

16 x^{2} + 25 y^{2} - 224 x + 250 y = 191

\frac{( x - 1 ) ^{2}}{40} + \frac{( y + 4 ) ^{2}}{36} = 1

\frac{x ^{2}}{( \frac{5}{2} ) ^{2}} + \frac{y ^{2}}{( \frac{3}{2} ) ^{2}} = 1

What is the foci ((x+2)^2)/(22)+((y+3)^2)/(18)=1 ?
The foci ((x+2)^2)/(22)+((y+3)^2)/(18)=1 is (0,-3),(-4,-3)