What is the foci (x^2)/5+(y^2)/4 =1 ?

The foci (x^2)/5+(y^2)/4 =1 is (1,0),(-1,0)

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foci (x^2)/5+(y^2)/4 =1

foci

\frac{x ^{2}}{5} + \frac{y ^{2}}{4} = 1

(1, 0), (- 1, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{5} + \frac{y ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 5, b = 2

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

Compute

c :

c = 5^{2} - 2^{2} : 1

(0 + 1, 0), (0 - 1, 0)

Refine

(1, 0), (- 1, 0)

\frac{( x + 20 ) ^{2}}{105} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{12} + \frac{( y ) ^{2}}{16} = 1

64 x^{2} + 100 y^{2} - 256 x + 200 y - 6044 = 0

\frac{( x + 10 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{5} = 1

16 x^{2} + 36 y^{2} = 576

What is the foci (x^2)/5+(y^2)/4 =1 ?
The foci (x^2)/5+(y^2)/4 =1 is (1,0),(-1,0)