What is the foci (x^2)/9+(y^2)/(16)>1 ?

The foci (x^2)/9+(y^2)/(16)>1 is (0,sqrt(7)),(0,-sqrt(7))

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foci (x^2)/9+(y^2)/(16)>1

foci

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

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

Solution steps

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

Rewrite as

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

Calculate ellipse properties

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

4 x^{2} + 9 y^{2} + 8 x - 36 y + 4 = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1

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

9 x^{2} + 16 y^{2} - 54 x + 64 y + 1 = 0

\frac{x ^{2}}{81} + \frac{y ^{2}}{16} = 1

What is the foci (x^2)/9+(y^2)/(16)>1 ?
The foci (x^2)/9+(y^2)/(16)>1 is (0,sqrt(7)),(0,-sqrt(7))