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

The foci (y^2}{16}+\frac{x^2)/9 =1 is (0,sqrt(7)),(0,-sqrt(7))

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

foci

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

(0, 7), (0, - 7)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

(0, 7), (0, - 7)

x^{2} + 3 y^{2} - 6 x + 6 y = 0

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

\frac{( x - 3 ) ^{2}}{16} + \frac{y ^{2}}{9} = 1

\frac{( x + 8 ) ^{2}}{169} + \frac{( y + 9 ) ^{2}}{25} = 1

9 x^{2} + 25 y^{2} + 36 x + 250 y = - 436

What is the foci (y^2}{16}+\frac{x^2)/9 =1 ?
The foci (y^2}{16}+\frac{x^2)/9 =1 is (0,sqrt(7)),(0,-sqrt(7))