What is the foci 1x^2+4y^2+10x+9y-10=0 ?

The foci 1x^2+4y^2+10x+9y-10=0 is ((-40+sqrt(1923))/8 ,-9/8),((-40-sqrt(1923))/8 ,-9/8)

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foci 1x^2+4y^2+10x+9y-10=0

foci

1 x^{2} + 4 y^{2} + 10 x + 9 y - 10 = 0

(\frac{- 40 + 1923}{8}, - \frac{9}{8}), (\frac{- 40 - 1923}{8}, - \frac{9}{8})

Solution steps

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

Calculate ellipse properties

1 x^{2} + 4 y^{2} + 10 x + 9 y - 10 = 0 :

Ellipse with center

(h, k) = (- 5, - \frac{9}{8}), a = \frac{641}{4}, b = \frac{641}{8}

(- 5 + c, - \frac{9}{8}), (- 5 - c, - \frac{9}{8})

Compute

c :

c = (\frac{641}{4})^{2} - (\frac{641}{8})^{2} : \frac{1923}{8}

(- 5 + \frac{1923}{8}, - \frac{9}{8}), (- 5 - \frac{1923}{8}, - \frac{9}{8})

Refine

(\frac{- 40 + 1923}{8}, - \frac{9}{8}), (\frac{- 40 - 1923}{8}, - \frac{9}{8})

\frac{( x + 2 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{25} = 1

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{100} = 1

x^{2} + 16 y^{2} - 64 = 0

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

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

What is the foci 1x^2+4y^2+10x+9y-10=0 ?
The foci 1x^2+4y^2+10x+9y-10=0 is ((-40+sqrt(1923))/8 ,-9/8),((-40-sqrt(1923))/8 ,-9/8)