SOLUTION: (1/2)x^2+(1/8)y^2=(1/4).... How do I graph this. I know I should get one on the left hand side. But then ill get 2x^2 +y^2/2=1 so from here how do I know the a and b if the 2x^2 is
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-> SOLUTION: (1/2)x^2+(1/8)y^2=(1/4).... How do I graph this. I know I should get one on the left hand side. But then ill get 2x^2 +y^2/2=1 so from here how do I know the a and b if the 2x^2 is
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Question 481921: (1/2)x^2+(1/8)y^2=(1/4).... How do I graph this. I know I should get one on the left hand side. But then ill get 2x^2 +y^2/2=1 so from here how do I know the a and b if the 2x^2 isn't a fraction w denominator... Found 2 solutions by ewatrrr, solver91311:Answer by ewatrrr(24785) (Show Source):
Hi,
Note:
Standard Form of an Equation of an Ellipse is
where Pt(h,k) is the center and a and b are the respective vertices distances from center.
In this case: a = sqrt(.5) and b = sqrt(2)
