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

The foci x^2+(y^2)/(16)=1 is (0,sqrt(15)),(0,-sqrt(15))

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

foci

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

(0, 15), (0, - 15)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 4, a = 1

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

Compute

c :

c = 4^{2} - 1^{2} : 15

(0, 0 + 15), (0, 0 - 15)

Refine

(0, 15), (0, - 15)

x^{2} = - 2 y

x = - 2 y^{2}

x^{2} = 20 y

3 x^{2} + y^{2} + 18 x - 2 y - 8 = 0

\frac{x ^{2}}{5} + \frac{y ^{2}}{9} = 1

What is the foci x^2+(y^2)/(16)=1 ?
The foci x^2+(y^2)/(16)=1 is (0,sqrt(15)),(0,-sqrt(15))