What is the foci (x^2)/(12)+((y-4)^2)/(16)=1 ?

The foci (x^2)/(12)+((y-4)^2)/(16)=1 is (0,6),(0,2)

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foci (x^2)/(12)+((y-4)^2)/(16)=1

foci

\frac{x ^{2}}{12} + \frac{( y - 4 ) ^{2}}{16} = 1

(0, 6), (0, 2)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{12} + \frac{( y - 4 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 4), b = 4, a = 23

(0, 4 + c), (0, 4 - c)

Compute

c :

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

(0, 4 + 2), (0, 4 - 2)

Refine

(0, 6), (0, 2)

\frac{( x + 2 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1

\frac{x ^{2}}{96} + \frac{y ^{2}}{100} = 1

9 x^{2} + 25 y^{2} - 100 y - 125 = 0

3 x^{2} + 4 y^{2} = 12

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{9} = 1

What is the foci (x^2)/(12)+((y-4)^2)/(16)=1 ?
The foci (x^2)/(12)+((y-4)^2)/(16)=1 is (0,6),(0,2)