What is the foci x^2+y^2+10x-6y+33=0 ?

The foci x^2+y^2+10x-6y+33=0 is (-5,3),(-5,3)

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foci x^2+y^2+10x-6y+33=0

foci

x^{2} + y^{2} + 10 x - 6 y + 33 = 0

(- 5, 3), (- 5, 3)

Solution steps

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

Calculate ellipse properties

x^{2} + y^{2} + 10 x - 6 y + 33 = 0 :

Ellipse with center

(h, k) = (- 5, 3), b = 1, a = 1

(- 5, 3 + c), (- 5, 3 - c)

Compute

c :

c = 1^{2} - 1^{2} : 0

(- 5, 3 + 0), (- 5, 3 - 0)

Refine

(- 5, 3), (- 5, 3)

\frac{( x - 2 ) ^{2}}{36} + \frac{( y - 6 ) ^{2}}{4} = 1

4 x^{2} + y^{2} - 48 x + 4 y + 48 = 0

7 x^{2} = 14 - 2 y^{2}

25 x^{2} + 4 y^{2} - 100 = 0

\frac{x ^{2}}{1.586 9 ^{2}} + \frac{y ^{2}}{0.9989} = 1

What is the foci x^2+y^2+10x-6y+33=0 ?
The foci x^2+y^2+10x-6y+33=0 is (-5,3),(-5,3)