What is the foci 2x^2-3y^2=6 ?

The foci 2x^2-3y^2=6 is (sqrt(5),0),(-sqrt(5),0)

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foci 2x^2-3y^2=6

foci

2 x^{2} - 3 y^{2} = 6

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

Solution steps

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

Calculate Hyperbola properties

2 x^{2} - 3 y^{2} = 6 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 2

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

Compute

c :

c = 3^{2} + 2^{2} : 5

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

Refine

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

49 y^{2} - x^{2} = 49

\frac{x ^{2}}{81} - \frac{y ^{2}}{1 6 ^{2}} = 1

9 x^{2} - 4 y^{2} + 72 x + 180 = 0

1 = x^{2} - y^{2}

\frac{x ^{2}}{900} - \frac{y ^{2}}{1600} = 1