What is the foci 9x^2-4y^2=1 ?

The foci 9x^2-4y^2=1 is ((sqrt(13))/6 ,0),(-(sqrt(13))/6 ,0)

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foci 9x^2-4y^2=1

foci

(\frac{13}{6}, 0), (- \frac{13}{6}, 0)

Solution steps

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

Calculate Hyperbola properties

9 x^{2} - 4 y^{2} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = \frac{1}{3}, b = \frac{1}{2}

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

Compute

c :

c = (\frac{1}{3})^{2} + (\frac{1}{2})^{2} : \frac{13}{6}

(0 + \frac{13}{6}, 0), (0 - \frac{13}{6}, 0)

Refine

(\frac{13}{6}, 0), (- \frac{13}{6}, 0)

What is the foci 9x^2-4y^2=1 ?
The foci 9x^2-4y^2=1 is ((sqrt(13))/6 ,0),(-(sqrt(13))/6 ,0)