What is the foci (x^2)/4+(y^2)/(16)=1 ?

The foci (x^2)/4+(y^2)/(16)=1 is (0,2sqrt(3)),(0,-2sqrt(3))

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foci (x^2)/4+(y^2)/(16)=1

foci

\frac{x ^{2}}{4} + \frac{y ^{2}}{16} = 1

(0, 23), (0, - 23)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 4, a = 2

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

Compute

c :

c = 4^{2} - 2^{2} : 23

(0, 0 + 23), (0, 0 - 23)

Refine

(0, 23), (0, - 23)

3 x^{2} + 2 y^{2} = 6

\frac{x ^{2}}{25} + \frac{y ^{2}}{4} = 1

4 x^{2} + y^{2} = 16

y^{2} = - 8 x

x = y^{2} - 5

What is the foci (x^2)/4+(y^2)/(16)=1 ?
The foci (x^2)/4+(y^2)/(16)=1 is (0,2sqrt(3)),(0,-2sqrt(3))