What is the foci 1x^2+3y^2+10x+8y-10=0 ?

The foci 1x^2+3y^2+10x+8y-10=0 is ((-15+11sqrt(2))/3 ,-4/3),((-15-11sqrt(2))/3 ,-4/3)

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foci 1x^2+3y^2+10x+8y-10=0

foci

1 x^{2} + 3 y^{2} + 10 x + 8 y - 10 = 0

(\frac{- 15 + 11 2}{3}, - \frac{4}{3}), (\frac{- 15 - 11 2}{3}, - \frac{4}{3})

Solution steps

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

Calculate ellipse properties

1 x^{2} + 3 y^{2} + 10 x + 8 y - 10 = 0 :

Ellipse with center

(h, k) = (- 5, - \frac{4}{3}), a = \frac{11}{3}, b = \frac{11}{3}

(- 5 + c, - \frac{4}{3}), (- 5 - c, - \frac{4}{3})

Compute

c :

c = (\frac{11}{3})^{2} - (\frac{11}{3})^{2} : \frac{11 2}{3}

(- 5 + \frac{11 2}{3}, - \frac{4}{3}), (- 5 - \frac{11 2}{3}, - \frac{4}{3})

Refine

(\frac{- 15 + 11 2}{3}, - \frac{4}{3}), (\frac{- 15 - 11 2}{3}, - \frac{4}{3})

4 x^{2} + y^{2} - 8 x + 4 y + 4 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{64} = 1

81 x^{2} + 4 y^{2} = 324

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 5 ) ^{2}}{9} = 1

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{49} = 1

What is the foci 1x^2+3y^2+10x+8y-10=0 ?
The foci 1x^2+3y^2+10x+8y-10=0 is ((-15+11sqrt(2))/3 ,-4/3),((-15-11sqrt(2))/3 ,-4/3)