What is the foci 8x^2+2y^2+16x=0 ?

The foci 8x^2+2y^2+16x=0 is (-1,sqrt(3)),(-1,-sqrt(3))

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foci 8x^2+2y^2+16x=0

foci

8 x^{2} + 2 y^{2} + 16 x = 0

(- 1, 3), (- 1, - 3)

Solution steps

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

Calculate ellipse properties

8 x^{2} + 2 y^{2} + 16 x = 0 :

Ellipse with center

(h, k) = (- 1, 0), b = 2, a = 1

(- 1, 0 + c), (- 1, 0 - c)

Compute

c :

c = 2^{2} - 1^{2} : 3

(- 1, 0 + 3), (- 1, 0 - 3)

Refine

(- 1, 3), (- 1, - 3)

\frac{x ^{2}}{169} + \frac{y ^{2}}{144} = 1

16 x^{2} + 25 y^{2} - 64 x - 50 y - 311 = 0

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{4} = 1

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

\frac{x ^{2}}{16} + \frac{y ^{2}}{81} = 1

What is the foci 8x^2+2y^2+16x=0 ?
The foci 8x^2+2y^2+16x=0 is (-1,sqrt(3)),(-1,-sqrt(3))