What is the foci x^2+4y^2+6x+16y+21=0 ?

The foci x^2+4y^2+6x+16y+21=0 is (-3+sqrt(3),-2),(-3-sqrt(3),-2)

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foci x^2+4y^2+6x+16y+21=0

foci

x^{2} + 4 y^{2} + 6 x + 16 y + 21 = 0

(- 3 + 3, - 2), (- 3 - 3, - 2)

Solution steps

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

Calculate ellipse properties

x^{2} + 4 y^{2} + 6 x + 16 y + 21 = 0 :

Ellipse with center

(h, k) = (- 3, - 2), a = 2, b = 1

(- 3 + c, - 2), (- 3 - c, - 2)

Compute

c :

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

(- 3 + 3, - 2), (- 3 - 3, - 2)

3 x^{2} + y^{2} = 6

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

\frac{( x ^{2} )}{49} + \frac{( y ^{2} )}{25} = 1

9 x^{2} + 4 y^{2} + 32 y + 28 = 0

\frac{x ^{2}}{625} + \frac{y ^{2}}{400} = 1

What is the foci x^2+4y^2+6x+16y+21=0 ?
The foci x^2+4y^2+6x+16y+21=0 is (-3+sqrt(3),-2),(-3-sqrt(3),-2)