What is the foci (y^2)/(81)+(x^2)/(36)=1 ?

The foci (y^2)/(81)+(x^2)/(36)=1 is (0,3sqrt(5)),(0,-3sqrt(5))

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foci (y^2)/(81)+(x^2)/(36)=1

foci

\frac{y ^{2}}{81} + \frac{x ^{2}}{36} = 1

(0, 35), (0, - 35)

Solution steps

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

Calculate ellipse properties

\frac{y ^{2}}{81} + \frac{x ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 9, a = 6

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

Compute

c :

c = 9^{2} - 6^{2} : 35

(0, 0 + 35), (0, 0 - 35)

Refine

(0, 35), (0, - 35)

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 1 ) ^{2}}{36} = 1

16 x^{2} + 25 y^{2} + 100 y - 300 = 0

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{4 ^{2}} = 1

x^{2} + 3 y^{2} = 1

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

What is the foci (y^2)/(81)+(x^2)/(36)=1 ?
The foci (y^2)/(81)+(x^2)/(36)=1 is (0,3sqrt(5)),(0,-3sqrt(5))