What is the foci 18x^2+24x+12y^2-12y=1 ?

The foci 18x^2+24x+12y^2-12y=1 is (-2/3 ,(1+2sqrt(1/3))/2),(-2/3 ,(1-2sqrt(1/3))/2)

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foci 18x^2+24x+12y^2-12y=1

foci

18 x^{2} + 24 x + 12 y^{2} - 12 y = 1

- \frac{2}{3}, \frac{1 + 2 \frac{1}{3}}{2}, - \frac{2}{3}, \frac{1 - 2 \frac{1}{3}}{2}

Solution steps

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

Calculate ellipse properties

18 x^{2} + 24 x + 12 y^{2} - 12 y = 1 :

Ellipse with center

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

(- \frac{2}{3}, \frac{1}{2} + c), (- \frac{2}{3}, \frac{1}{2} - c)

Compute

c :

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

(- \frac{2}{3}, \frac{1}{2} + \frac{1}{3}), (- \frac{2}{3}, \frac{1}{2} - \frac{1}{3})

Refine

- \frac{2}{3}, \frac{1 + 2 \frac{1}{3}}{2}, - \frac{2}{3}, \frac{1 - 2 \frac{1}{3}}{2}

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

x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0

x^{2} + 3 y^{2} - 4 x + 12 y + 13 = 0

49 x^{2} + 25 y^{2} = 1225

\frac{x ^{2}}{64} + \frac{y ^{2}}{49} = 1

What is the foci 18x^2+24x+12y^2-12y=1 ?
The foci 18x^2+24x+12y^2-12y=1 is (-2/3 ,(1+2sqrt(1/3))/2),(-2/3 ,(1-2sqrt(1/3))/2)