What is the foci ((x)^2}{13}+\frac{y^2)/4 =1 ?

The foci ((x)^2}{13}+\frac{y^2)/4 =1 is (3,0),(-3,0)

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foci ((x)^2}{13}+\frac{y^2)/4 =1

foci

\frac{( x ) ^{2}}{13} + \frac{y ^{2}}{4} = 1

(3, 0), (- 3, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{13} + \frac{y ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 13, b = 2

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

Compute

c :

c = 13^{2} - 2^{2} : 3

(0 + 3, 0), (0 - 3, 0)

Refine

(3, 0), (- 3, 0)

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{16} = 1

\frac{1 x ^{2}}{16} + \frac{y ^{2}}{9} = 1

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{49} = 1

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

What is the foci ((x)^2}{13}+\frac{y^2)/4 =1 ?
The foci ((x)^2}{13}+\frac{y^2)/4 =1 is (3,0),(-3,0)