What is the vertices x^2+3y^2+2x-12y+10=0 ?

The vertices x^2+3y^2+2x-12y+10=0 is (-1+sqrt(3),2),(-1-sqrt(3),2)

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vertices x^2+3y^2+2x-12y+10=0

vertices

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0

(- 1 + 3, 2), (- 1 - 3, 2)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0 :

Ellipse with center

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

(- 1 + 3, 2), (- 1 - 3, 2)

9 x^{2} + 16 y^{2} = 576

\frac{x ^{2}}{8} + \frac{y ^{2}}{5} = 1

9 x^{2} + 16 y^{2} - 144 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1

16 x^{2} + 9 y^{2} - 64 x - 54 y + 1 = 0

What is the vertices x^2+3y^2+2x-12y+10=0 ?
The vertices x^2+3y^2+2x-12y+10=0 is (-1+sqrt(3),2),(-1-sqrt(3),2)