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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

- \frac{4}{3} + 2 x + x^{2} + \frac{4 y}{3} + \frac{y ^{2}}{3} = 0

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{4} = 1

\frac{( x - 1 ) ^{2}}{8} + \frac{( y - 3 ) ^{2}}{4} = 1

\frac{x ^{2}}{4} + \frac{y ^{2}}{20} = 1

\frac{( x 8 ) ^{2}}{49} + \frac{( y + 1 ) ^{2}}{100} = 1

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