SOLUTION: Graph, find center, vertices, co-vertices,foci, and eccentricity e. 9x^2+4y^2=36

Question 190926This question is from textbook saxon algebra 2
: Graph, find center, vertices, co-vertices,foci, and eccentricity e.
9x^2+4y^2=36 This question is from textbook saxon algebra 2

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Graph, find center, vertices, co-vertices,foci, and eccentricity e.
9x^2+4y^2=36
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Standard form is:
x^2/4 + y^2/9 = 1
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center is (0,0)
vertices: (0,3) and (0,-3)
co-vertices: (2,0) and (-2,0)
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a = 3, b = 2
For an ellipse a^2 = b^2 + c^2
c = sqrt(a^2-b^2) = sqrt(9-4) = sqrt(5)
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Foci: (0,sqrt(5)) , (0,-sqrt(5))
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e = c/a = sqrt(5)/3
=========================================
Cheers,
Stan H.
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