What is the foci 4x^2+8y^2=3 ?

The foci 4x^2+8y^2=3 is ((sqrt(3))/(2sqrt(2)),0),(-(sqrt(3))/(2sqrt(2)),0)

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foci 4x^2+8y^2=3

foci

4 x^{2} + 8 y^{2} = 3

(\frac{3}{2 2}, 0), (- \frac{3}{2 2}, 0)

Solution steps

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

Calculate ellipse properties

4 x^{2} + 8 y^{2} = 3 :

Ellipse with center

(h, k) = (0, 0), a = \frac{3}{2}, b = \frac{3}{2 2}

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

Compute

c :

c = (\frac{3}{2})^{2} - (\frac{3}{2 2})^{2} : \frac{3}{2 2}

(0 + \frac{3}{2 2}, 0), (0 - \frac{3}{2 2}, 0)

Refine

(\frac{3}{2 2}, 0), (- \frac{3}{2 2}, 0)

16 x^{2} + 4 y^{2} - 64 = 0

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{( x - 4 ) ^{2}}{3} + \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{36} + \frac{8 y ^{2}}{81} = 1

x^{2} + 3 y^{2} + 8 x + 9 y - 14 = 0

What is the foci 4x^2+8y^2=3 ?
The foci 4x^2+8y^2=3 is ((sqrt(3))/(2sqrt(2)),0),(-(sqrt(3))/(2sqrt(2)),0)