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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

x^{2} - 4 y = 0

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

x^{2} = 4 y - 2 y^{2}

x = (y - 5)^{2}

(x - 2)^{2} = 12 (y + 3)

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