What is the foci ((x+4)^2)/(22)+((y+3)^2)/(26)=1 ?

The foci ((x+4)^2)/(22)+((y+3)^2)/(26)=1 is (-4,-1),(-4,-5)

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foci ((x+4)^2)/(22)+((y+3)^2)/(26)=1

foci

\frac{( x + 4 ) ^{2}}{22} + \frac{( y + 3 ) ^{2}}{26} = 1

(- 4, - 1), (- 4, - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 4 ) ^{2}}{22} + \frac{( y + 3 ) ^{2}}{26} = 1 :

Ellipse with center

(h, k) = (- 4, - 3), b = 26, a = 22

(- 4, - 3 + c), (- 4, - 3 - c)

Compute

c :

c = 26^{2} - 22^{2} : 2

(- 4, - 3 + 2), (- 4, - 3 - 2)

Refine

(- 4, - 1), (- 4, - 5)

+ 4 x^{2} + 9 y^{2} - 16 x + 18 y = 11

16 (x - 4)^{2} + 25 (y + 2)^{2} = 400

\frac{x ^{2}}{38} + \frac{y ^{2}}{22} = 1

\frac{( x - 4 ) ^{2}}{25} + \frac{( y - 2 ) ^{2}}{16} = 1

14 x^{2} + 5 y^{2} - 70 = 0

What is the foci ((x+4)^2)/(22)+((y+3)^2)/(26)=1 ?
The foci ((x+4)^2)/(22)+((y+3)^2)/(26)=1 is (-4,-1),(-4,-5)