What is the foci (x^2}{16}+\frac{y^2)/4 =1 ?

The foci (x^2}{16}+\frac{y^2)/4 =1 is (2sqrt(3),0),(-2sqrt(3),0)

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foci (x^2}{16}+\frac{y^2)/4 =1

foci

\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

(23, 0), (- 23, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 4, b = 2

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

Compute

c :

c = 4^{2} - 2^{2} : 23

(0 + 23, 0), (0 - 23, 0)

Refine

(23, 0), (- 23, 0)

x^{2} + 2 y - 6 = 0

4 y^{2} - 48 x - 20 y = 71

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

f (x) = x^{2} - 6 x + 8

x^{2} - y^{2} = 10 (x - y) + 1

What is the foci (x^2}{16}+\frac{y^2)/4 =1 ?
The foci (x^2}{16}+\frac{y^2)/4 =1 is (2sqrt(3),0),(-2sqrt(3),0)