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

The foci (x^2)/(16)+(y-4)^2=1 is (sqrt(15),4),(-sqrt(15),4)

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

foci

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

(15, 4), (- 15, 4)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

(0 + 15, 4), (0 - 15, 4)

Refine

(15, 4), (- 15, 4)

3 x^{2} + y^{2} - 24 x + 39 = 0

9 x^{2} - 18 x + 4 y^{2} = 27

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

4 x^{2} + 9 y^{2} - 16 x + 18 y - 11 = 0

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

What is the foci (x^2)/(16)+(y-4)^2=1 ?
The foci (x^2)/(16)+(y-4)^2=1 is (sqrt(15),4),(-sqrt(15),4)