What is the foci (x^2}{16}+\frac{(y-2)^2)/9 =1 ?

The foci (x^2}{16}+\frac{(y-2)^2)/9 =1 is (sqrt(7),2),(-sqrt(7),2)

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

foci

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

(7, 2), (- 7, 2)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(0 + c, 2), (0 - c, 2)

Compute

c :

c = 4^{2} - 3^{2} : 7

(0 + 7, 2), (0 - 7, 2)

Refine

(7, 2), (- 7, 2)

\frac{( x + 4 ) ^{2}}{225} + \frac{( y - 7 ) ^{2}}{81} = 1

9 x^{2} + 4 y^{2} = 24 y

25 x^{2} + 16 y^{2} + 100 x - 96 y - 156 = 0

2 x^{2} + 4 y^{2} - 8 x - 4 y - 15 = 0

3 x^{2} + y^{2} = 9

What is the foci (x^2}{16}+\frac{(y-2)^2)/9 =1 ?
The foci (x^2}{16}+\frac{(y-2)^2)/9 =1 is (sqrt(7),2),(-sqrt(7),2)