What is the foci x^2+4y^2+10x+8y-3=0 ?

The foci x^2+4y^2+10x+8y-3=0 is (-5+2sqrt(6),-1),(-5-2sqrt(6),-1)

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foci x^2+4y^2+10x+8y-3=0

foci

x^{2} + 4 y^{2} + 10 x + 8 y - 3 = 0

(- 5 + 26, - 1), (- 5 - 26, - 1)

Solution steps

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

Calculate ellipse properties

x^{2} + 4 y^{2} + 10 x + 8 y - 3 = 0 :

Ellipse with center

(h, k) = (- 5, - 1), a = 42, b = 22

(- 5 + c, - 1), (- 5 - c, - 1)

Compute

c :

c = (42)^{2} - (22)^{2} : 26

(- 5 + 26, - 1), (- 5 - 26, - 1)

36 x^{2} + y^{2} = 36

\frac{( x + 1 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{25} = 1

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

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

x^{2} + 2 y^{2} - 2 x + 8 y + 5 = 0

What is the foci x^2+4y^2+10x+8y-3=0 ?
The foci x^2+4y^2+10x+8y-3=0 is (-5+2sqrt(6),-1),(-5-2sqrt(6),-1)