What is the foci (x^2)/5+(y^2)/(16)=1 ?

The foci (x^2)/5+(y^2)/(16)=1 is (0,sqrt(11)),(0,-sqrt(11))

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foci (x^2)/5+(y^2)/(16)=1

foci

\frac{x ^{2}}{5} + \frac{y ^{2}}{16} = 1

(0, 11), (0, - 11)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{5} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 4, a = 5

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

Compute

c :

c = 4^{2} - 5^{2} : 11

(0, 0 + 11), (0, 0 - 11)

Refine

(0, 11), (0, - 11)

\frac{x ^{2}}{16} + \frac{y ^{2}}{8} = 1

9 x^{2} + 4 y^{2} - 54 x + 40 y + 145 = 0

9 x^{2} + y^{2} + 36 x - 2 y + 1 = 0

\frac{( x ) ^{2}}{46} + \frac{( y + 8 ) ^{2}}{26} = 1

\frac{( x - 4 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{25} = 1

What is the foci (x^2)/5+(y^2)/(16)=1 ?
The foci (x^2)/5+(y^2)/(16)=1 is (0,sqrt(11)),(0,-sqrt(11))