What is the foci ((x-2)^2)/9+((y+1)^2)/(36)=1 ?

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

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foci ((x-2)^2)/9+((y+1)^2)/(36)=1

foci

(2, - 1 + 33), (2, - 1 - 33)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, - 1), b = 6, a = 3

(2, - 1 + c), (2, - 1 - c)

Compute

c :

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

(2, - 1 + 33), (2, - 1 - 33)

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