What is the foci x^2+((y-6)^2)/(16)=1 ?

The foci x^2+((y-6)^2)/(16)=1 is (0,6+sqrt(15)),(0,6-sqrt(15))

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foci x^2+((y-6)^2)/(16)=1

foci

x^{2} + \frac{( y - 6 ) ^{2}}{16} = 1

(0, 6 + 15), (0, 6 - 15)

Solution steps

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

Calculate ellipse properties

x^{2} + \frac{( y - 6 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 6), b = 4, a = 1

(0, 6 + c), (0, 6 - c)

Compute

c :

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

(0, 6 + 15), (0, 6 - 15)

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

\frac{( x + 5 ) ^{2}}{20} + \frac{( y - 6 ) ^{2}}{11} = 1

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 3

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

x^{2} + 4 y^{2} - 4 x - 8 y - 92 = 0

What is the foci x^2+((y-6)^2)/(16)=1 ?
The foci x^2+((y-6)^2)/(16)=1 is (0,6+sqrt(15)),(0,6-sqrt(15))