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)

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)

\frac{x ^{2}}{36} - \frac{y ^{2}}{64} = 1

x^{2} = - 6 y

y^{2} - 4 y - 4 x = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1

\frac{x ^{2}}{25} + \frac{y ^{2}}{25} = 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))