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

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

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foci (x^2}{16}+\frac{y^2)/5 =1

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

81 x^{2} + 16 y^{2} = 1296

25 x^{2} + 9 y^{2} - 100 x - 90 y + 100 = 0

5 x^{2} + 7 y^{2} - 35 = 0

4 (x - 1)^{2} + 13 (y + 3)^{2} = 52

x^{2} + 5 y^{2} - 4 x + 20 y = - 19

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