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

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

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foci (x^2)/4+((y-3)^2)/(16)=1

foci

\frac{x ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

(0, 3 + 23), (0, 3 - 23)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 3), b = 4, a = 2

(0, 3 + c), (0, 3 - c)

Compute

c :

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

(0, 3 + 23), (0, 3 - 23)

\frac{x ^{2}}{9} + \frac{y ^{2}}{81} = 1

16 xx + 25 yy = 400

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

16 x^{2} + 25 y^{2} - 64 x - 336 = 0

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

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