What is the foci ((x+5)^2)/(20)+((y-6)^2)/(11)=1 ?

The foci ((x+5)^2)/(20)+((y-6)^2)/(11)=1 is (-2,6),(-8,6)

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foci ((x+5)^2)/(20)+((y-6)^2)/(11)=1

foci

\frac{( x + 5 ) ^{2}}{20} + \frac{( y - 6 ) ^{2}}{11} = 1

(- 2, 6), (- 8, 6)

Solution steps

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

Calculate ellipse properties

\frac{( x + 5 ) ^{2}}{20} + \frac{( y - 6 ) ^{2}}{11} = 1 :

Ellipse with center

(h, k) = (- 5, 6), a = 25, b = 11

(- 5 + c, 6), (- 5 - c, 6)

Compute

c :

c = (25)^{2} - 11^{2} : 3

(- 5 + 3, 6), (- 5 - 3, 6)

Refine

(- 2, 6), (- 8, 6)

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 3

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

x^{2} + 4 y^{2} - 4 x - 8 y - 92 = 0

x^{2} + 9 y^{2} = 36

9 x^{2} + 25 y^{2} + 108 x - 200 y + 499 = 0

What is the foci ((x+5)^2)/(20)+((y-6)^2)/(11)=1 ?
The foci ((x+5)^2)/(20)+((y-6)^2)/(11)=1 is (-2,6),(-8,6)