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

The foci 4x^2+9y^2=1 is ((sqrt(5))/6 ,0),(-(sqrt(5))/6 ,0)

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

foci

4 x^{2} + 9 y^{2} = 1

(\frac{5}{6}, 0), (- \frac{5}{6}, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

c = (\frac{1}{2})^{2} - (\frac{1}{3})^{2} : \frac{5}{6}

(0 + \frac{5}{6}, 0), (0 - \frac{5}{6}, 0)

Refine

(\frac{5}{6}, 0), (- \frac{5}{6}, 0)

25 x^{2} + 16 y^{2} - 100 x - 300 = 0

\frac{( x - 2 ) ^{2}}{2} + \frac{( y - 1 ) ^{2}}{8} = 1

9 x^{2} + 4 y^{2} = 25

4 x^{2} + 9 y^{2} + 24 x - 36 y + 36 = 0

4 x^{2} + 16 y^{2} - 16 x - 48 = 0

What is the foci 4x^2+9y^2=1 ?
The foci 4x^2+9y^2=1 is ((sqrt(5))/6 ,0),(-(sqrt(5))/6 ,0)