“Find the center, vertices, and foci of the ellipse with equation 2×2 + 6y2 = 12.?

2x^2 + 6y^2 = 12, ie., (x/rt6)^2 + (y/rt2)^2 = 1, ie., (x/a)^2 + (y/b)^2 = 1, where a = rt6, b = rt2. Center = (0,0),

major vertices are ([+/-]a,0) = ([+/-]rt6,0), minor vertices are (0,[+/-]b) = (0,[+/-]rt2), foci are ([+/-]c,0) = ([+/-]2,0).

Note: c^2 = a^2 – b^2 = (rt6)^2 – (rt2)^2 = 6-2 = 4 & c = 2.
You mean this?

2x² + 6y² = 12

Try writing it in standard form. That helps show many of the parameters.

2x² + 6y² = 12

x²/6 + y²/2 = 1

(x – 0)²/√(6)² + (y – 0)²/√(2)² = 1

center: (0, 0)

semimajor (horizontal): √(6)

semiminor (vertical): √(2)

focal distance = √(6 – 2) = 2

major vertices: (√(6), 0), (-√(6), 0)

minor vertices: (0, √(2)), (0, -√(2))

foci: (2, 0), (-2, 0)