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

The foci 2x^2+3y^2+10x+10y-10=0 is ((-15+sqrt(185))/6 ,-5/3),((-15-sqrt(185))/6 ,-5/3)

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

foci

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

(\frac{- 15 + 185}{6}, - \frac{5}{3}), (\frac{- 15 - 185}{6}, - \frac{5}{3})

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- \frac{5}{2}, - \frac{5}{3}), a = \frac{185}{2 3}, b = \frac{185}{3 2}

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

Compute

c :

c = (\frac{185}{2 3})^{2} - (\frac{185}{3 2})^{2} : \frac{185}{6}

(- \frac{5}{2} + \frac{185}{6}, - \frac{5}{3}), (- \frac{5}{2} - \frac{185}{6}, - \frac{5}{3})

Refine

(\frac{- 15 + 185}{6}, - \frac{5}{3}), (\frac{- 15 - 185}{6}, - \frac{5}{3})

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{4} = 1

\frac{x ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

\frac{( x + 4 ) ^{2}}{225} + \frac{( y - 7 ) ^{2}}{81} = 1

9 x^{2} + 4 y^{2} = 24 y

25 x^{2} + 16 y^{2} + 100 x - 96 y - 156 = 0

What is the foci 2x^2+3y^2+10x+10y-10=0 ?
The foci 2x^2+3y^2+10x+10y-10=0 is ((-15+sqrt(185))/6 ,-5/3),((-15-sqrt(185))/6 ,-5/3)