What is the foci (x^2)/(20)+(y^2)/(36)=1 ?

The foci (x^2)/(20)+(y^2)/(36)=1 is (0,4),(0,-4)

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foci (x^2)/(20)+(y^2)/(36)=1

foci

\frac{x ^{2}}{20} + \frac{y ^{2}}{36} = 1

(0, 4), (0, - 4)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{20} + \frac{y ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 6, a = 25

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

Compute

c :

c = 6^{2} - (25)^{2} : 4

(0, 0 + 4), (0, 0 - 4)

Refine

(0, 4), (0, - 4)

\frac{x ^{2}}{12} + \frac{y ^{2}}{6} = 1

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{100} = 1

x^{2} + 4 y^{2} - 6 x + 16 y + 21 = 0

What is the foci (x^2)/(20)+(y^2)/(36)=1 ?
The foci (x^2)/(20)+(y^2)/(36)=1 is (0,4),(0,-4)