What is the foci (x^2)/(20)+(y^2)/(16)=1 ?

The foci (x^2)/(20)+(y^2)/(16)=1 is (2,0),(-2,0)

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foci (x^2)/(20)+(y^2)/(16)=1

foci

\frac{x ^{2}}{20} + \frac{y ^{2}}{16} = 1

(2, 0), (- 2, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{20} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 25, b = 4

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

Compute

c :

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

(0 + 2, 0), (0 - 2, 0)

Refine

(2, 0), (- 2, 0)

\frac{y ^{2}}{16} + \frac{x ^{2}}{25} = 1

\frac{1}{4} (x - 5)^{2} + \frac{1}{9} (y + 5)^{2} = 1

\frac{( x + 1 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{49} = 1

4 x^{2} + 36 y^{2} = 144

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

What is the foci (x^2)/(20)+(y^2)/(16)=1 ?
The foci (x^2)/(20)+(y^2)/(16)=1 is (2,0),(-2,0)