What is the vertices 2x^2+2y^2-28x+12y+114=0 ?

The vertices 2x^2+2y^2-28x+12y+114=0 is (7,-2),(7,-4)

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vertices 2x^2+2y^2-28x+12y+114=0

vertices

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0

(7, - 2), (7, - 4)

Solution steps

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

Calculate ellipse properties

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0 :

Ellipse with center

(h, k) = (7, - 3), b = 1, a = 1

(7, - 3 + 1), (7, - 3 - 1)

Refine

(7, - 2), (7, - 4)

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

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

x^{2} + y^{2} - 10 y = 0

25 x^{2} + 4 y^{2} + 200 x - 8 y + 304 = 0

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

What is the vertices 2x^2+2y^2-28x+12y+114=0 ?
The vertices 2x^2+2y^2-28x+12y+114=0 is (7,-2),(7,-4)