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

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

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(- 1 + 2, 1), (- 1 - 2, 1)

Refine

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

4 x^{2} + 16 y^{2} - 64 = 0

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

25 x^{2} + 9 y^{2} + 50 x - 108 y + 124 = 0

25 x^{2} + 4 y^{2} - 250 x - 16 y + 541 = 0

3 x^{2} + 4 y^{2} = 12

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