What is the vertices 2x^2+y^2+12x-6y+26=0 ?

The vertices 2x^2+y^2+12x-6y+26=0 is (-3,4),(-3,2)

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vertices 2x^2+y^2+12x-6y+26=0

vertices

2 x^{2} + y^{2} + 12 x - 6 y + 26 = 0

(- 3, 4), (- 3, 2)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

2 x^{2} + y^{2} + 12 x - 6 y + 26 = 0 :

Ellipse with center

(h, k) = (- 3, 3), b = 1, a = \frac{1}{2}

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

Refine

(- 3, 4), (- 3, 2)

\frac{x ^{2}}{36} + \frac{y ^{2}}{4} = 1

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

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

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

5 x^{2} + 2 y^{2} = 50

What is the vertices 2x^2+y^2+12x-6y+26=0 ?
The vertices 2x^2+y^2+12x-6y+26=0 is (-3,4),(-3,2)