What is the vertices 2x^2+3y^2+4x-6y-1=0 ?

The vertices 2x^2+3y^2+4x-6y-1=0 is (-1+sqrt(3),1),(-1-sqrt(3),1)

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vertices 2x^2+3y^2+4x-6y-1=0

vertices

2 x^{2} + 3 y^{2} + 4 x - 6 y - 1 = 0

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

Solution steps

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

Calculate ellipse properties

2 x^{2} + 3 y^{2} + 4 x - 6 y - 1 = 0 :

Ellipse with center

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

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

\frac{x ^{2}}{36} + \frac{y ^{2}}{25} = 1

36 x^{2} + 9 y^{2} = 324

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

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

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

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