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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

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

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

\frac{( x - 1 ) ^{2}}{6} + \frac{( y + 2 ) ^{2}}{18} = 1

\frac{( x + 11 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{1 9 ^{2}} = 1

\frac{( x + 6 ) ^{2}}{18} + \frac{( y + 3 ) ^{2}}{9} = 1

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