What is the foci (x^2)/4+(y^2)/(12)=1 ?

The foci (x^2)/4+(y^2)/(12)=1 is (0,2sqrt(2)),(0,-2sqrt(2))

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foci (x^2)/4+(y^2)/(12)=1

foci

\frac{x ^{2}}{4} + \frac{y ^{2}}{12} = 1

(0, 22), (0, - 22)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{y ^{2}}{12} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 23, a = 2

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

Compute

c :

c = (23)^{2} - 2^{2} : 22

(0, 0 + 22), (0, 0 - 22)

Refine

(0, 22), (0, - 22)

\frac{( x + 0 ) ^{2}}{100} + \frac{( y + 0 ) ^{2}}{64} = 1

2205 = 180 x^{2} + 196 y^{2}

\frac{x ^{2}}{0.38 1 ^{2}} + \frac{y ^{2}}{0.372 8 ^{2}} = 1

25 x^{2} + 9 y^{2} + 150 x = 0

4 x^{2} + y^{2} - 4 y = 0

What is the foci (x^2)/4+(y^2)/(12)=1 ?
The foci (x^2)/4+(y^2)/(12)=1 is (0,2sqrt(2)),(0,-2sqrt(2))