What is the foci (x^2)/7+(y^2)/(16)=1 ?

The foci (x^2)/7+(y^2)/(16)=1 is (0,3),(0,-3)

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foci (x^2)/7+(y^2)/(16)=1

foci

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1

(0, 3), (0, - 3)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 4, a = 7

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

Compute

c :

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

(0, 0 + 3), (0, 0 - 3)

Refine

(0, 3), (0, - 3)

y^{2} = \frac{7}{2} x

x^{2} = - 8 y

y^{2} = 3 x

\frac{x ^{2}}{4} - y^{2} = 1

\frac{x ^{2}}{25} + \frac{y ^{2}}{4} = 1

What is the foci (x^2)/7+(y^2)/(16)=1 ?
The foci (x^2)/7+(y^2)/(16)=1 is (0,3),(0,-3)