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

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

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vertices 4x^2+4x+y^2=0

vertices

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

(- \frac{1}{2}, 1), (- \frac{1}{2}, - 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

(- \frac{1}{2}, 1), (- \frac{1}{2}, - 1)

\frac{( x ^{2} )}{7} + \frac{( y ^{2} )}{16} = 1

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{10} = 1

\frac{x ^{2}}{9} + \frac{( y - 4 ) ^{2}}{7} = 1

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

\frac{( x - 1 ) ^{2}}{3} + \frac{( y - 1 ) ^{2}}{6} = 4

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