What is the foci (x^2)/(27)+(y^2)/(36)=1 ?

The foci (x^2)/(27)+(y^2)/(36)=1 is (0,3),(0,-3)

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foci (x^2)/(27)+(y^2)/(36)=1

foci

\frac{x ^{2}}{27} + \frac{y ^{2}}{36} = 1

(0, 3), (0, - 3)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{27} + \frac{y ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 6, a = 33

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

Compute

c :

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

(0, 0 + 3), (0, 0 - 3)

Refine

(0, 3), (0, - 3)

x^{2} + 4 x + 2 y^{2} + 6 y + 1 = 0

25 x^{2} + 81 y^{2} = 2025

x^{2} + 9 y^{2} - 14 x + 36 y + 49 = 0

4 x^{2} - 24 x + \frac{y ^{2}}{16} + \frac{y}{8} + \frac{573}{16} = 0

\frac{( x + 5 ) ^{2}}{5} + \frac{( y + 9 ) ^{2}}{9} = 1

What is the foci (x^2)/(27)+(y^2)/(36)=1 ?
The foci (x^2)/(27)+(y^2)/(36)=1 is (0,3),(0,-3)