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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{121} = 1

\frac{( x + 7 ) ^{2}}{8} + \frac{( y - 10 ) ^{2}}{13} = 1

\frac{( x - 1 ) ^{2}}{49} + \frac{( y - 3 ) ^{2}}{36} = 1

4 x^{2} + y^{2} - 32 x + 16 y + 124 = 0

1 x^{2} + 3 y^{2} + 10 x + 8 y - 10 = 0

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