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Find the center, vertices, and foci of the ellipse with equation

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2x^2 + 6y^2 = 12

asked Jul 3, 2013 in PRECALCULUS by angel12 Scholar

1 Answer

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The given ellipse equation is 2x^2+6y^2 = 12

The standard form of the equation of an ellipse is ( x - h)^2/a^2+( y - k)^2/b^2 = 1

2x^2/12+ 6y^2/12 =1

( x - 0)^2/6+( y - 0)^2/2 = 1

a^2 = 6 and  b^2 = 2

c^2 = a^2 - b^2

c^2 = 6 - 2 = 4

c^2 = 4

Therefore c = 2

a = √6, b =√2 and c = 2

The center at ( h, k) ⇒( 0,0)

vertices ( h, k±b)⇒ ( 0, 0±√2)

vertices at ( 0,√2) and ( 0, -√2)

foci ( h, k±c) ⇒( 0, 0± 2)

foci at ( 0, 2) and ( 0, -2)

The solution is center ( 0, 0) ,vertices ( 0,√2)& ( 0,-√2) and foci  ( 0, 2)& ( 0, -2).

answered Jul 3, 2013 by goushi Pupil

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