What is the foci (x^2)/8+(y^2)/3 =1 ?

The foci (x^2)/8+(y^2)/3 =1 is (sqrt(5),0),(-sqrt(5),0)

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foci (x^2)/8+(y^2)/3 =1

foci

\frac{x ^{2}}{8} + \frac{y ^{2}}{3} = 1

(5, 0), (- 5, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{8} + \frac{y ^{2}}{3} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 22, b = 3

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

Compute

c :

c = (22)^{2} - 3^{2} : 5

(0 + 5, 0), (0 - 5, 0)

Refine

(5, 0), (- 5, 0)

f (x) = x^{2} + 6 x + 5

4 x^{2} - 12 y^{2} - 16 x - 72 y - 44 = 0

y^{2} = - 2 x

(x - 2)^{2} + (y - 3)^{2} = 64

4 x^{2} - 9 y^{2} = 36

What is the foci (x^2)/8+(y^2)/3 =1 ?
The foci (x^2)/8+(y^2)/3 =1 is (sqrt(5),0),(-sqrt(5),0)