What is the vertices x^2+y^2+2x=3y ?

The vertices x^2+y^2+2x=3y is (-1,(3+sqrt(13))/2),(-1,(3-sqrt(13))/2)

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vertices x^2+y^2+2x=3y

vertices

x^{2} + y^{2} + 2 x = 3 y

(- 1, \frac{3 + 13}{2}), (- 1, \frac{3 - 13}{2})

Solution steps

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

Calculate ellipse properties

x^{2} + y^{2} + 2 x = 3 y :

Ellipse with center

(h, k) = (- 1, \frac{3}{2}), b = \frac{13}{2}, a = \frac{13}{2}

(- 1, \frac{3}{2} + \frac{13}{2}), (- 1, \frac{3}{2} - \frac{13}{2})

Refine

(- 1, \frac{3 + 13}{2}), (- 1, \frac{3 - 13}{2})

\frac{x ^{2}}{49} + \frac{y ^{2}}{169} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{64} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

9 x^{2} + y^{2} = 84

\frac{( x - 4 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1

What is the vertices x^2+y^2+2x=3y ?
The vertices x^2+y^2+2x=3y is (-1,(3+sqrt(13))/2),(-1,(3-sqrt(13))/2)