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

Algebra ->  Quadratic-relations-and-conic-sections -> 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      Log On

 

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)