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

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

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

vertices

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

(- 1 + \frac{3}{2}, 0), (- 1 - \frac{3}{2}, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 1, 0), a = \frac{3}{2}, b = 1

(- 1 + \frac{3}{2}, 0), (- 1 - \frac{3}{2}, 0)

25 x^{2} + 9 y^{2} + 200 x - 90 y + 400 = 0

9 x^{2} + 36 y^{2} + 36 x - 216 y + 36 = 0

9 x^{2} + 6 y^{2} + 36 y = 0

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

x^{2} + 4 y^{2} - 6 x - 16 y = 36

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