What is the vertices ((x+1)^2)/3+((y-1)^2)/6 =4 ?

The vertices ((x+1)^2)/3+((y-1)^2)/6 =4 is (-1,1+2sqrt(6)),(-1,1-2sqrt(6))

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vertices ((x+1)^2)/3+((y-1)^2)/6 =4

vertices

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

(- 1, 1 + 26), (- 1, 1 - 26)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 1, 1), b = 26, a = 23

(- 1, 1 + 26), (- 1, 1 - 26)

(x + 3)^{2} + y^{2} = 36

16 x^{2} + 25 y^{2} - 64 x + 150 y - 111 = 0

9 x^{2} + 16 y^{2} - 36 x + 96 y + 36 = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

What is the vertices ((x+1)^2)/3+((y-1)^2)/6 =4 ?
The vertices ((x+1)^2)/3+((y-1)^2)/6 =4 is (-1,1+2sqrt(6)),(-1,1-2sqrt(6))