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

The foci (x^2)/9+(y^2)/(36)=1 is (0,3sqrt(3)),(0,-3sqrt(3))

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

foci

\frac{x ^{2}}{9} + \frac{y ^{2}}{36} = 1

(0, 33), (0, - 33)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 6, a = 3

(0, 0 + c), (0, 0 - c)

Compute

c :

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

(0, 0 + 33), (0, 0 - 33)

Refine

(0, 33), (0, - 33)

4 (x - 1)^{2} + 9 (y - 1)^{2} = 36

\frac{( x + 1 ) ^{2}}{100} + \frac{( y + 2 ) ^{2}}{121} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{4} = 1

\frac{( x - 4 ) ^{2}}{24} + \frac{( y - 2 ) ^{2}}{49} = 1

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

What is the foci (x^2)/9+(y^2)/(36)=1 ?
The foci (x^2)/9+(y^2)/(36)=1 is (0,3sqrt(3)),(0,-3sqrt(3))