What is the foci ((x+3)^2}{36}+\frac{(y-1)^2)/9 =1 ?

The foci ((x+3)^2}{36}+\frac{(y-1)^2)/9 =1 is (-3+3sqrt(3),1),(-3-3sqrt(3),1)

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foci ((x+3)^2}{36}+\frac{(y-1)^2)/9 =1

foci

(- 3 + 33, 1), (- 3 - 33, 1)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 1 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (- 3, 1), a = 6, b = 3

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

Compute

c :

c = 6^{2} - 3^{2} : 33

(- 3 + 33, 1), (- 3 - 33, 1)

What is the foci ((x+3)^2}{36}+\frac{(y-1)^2)/9 =1 ?
The foci ((x+3)^2}{36}+\frac{(y-1)^2)/9 =1 is (-3+3sqrt(3),1),(-3-3sqrt(3),1)