SOLUTION: ellipse problem #11 find vertices, foci 2x^2+y^2=3 I can't figure out how to put in (x^2/a^2)+(y^2/b^2)=1 form. How do I get rid of the 2 from 2x? and make 3 into 1?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: ellipse problem #11 find vertices, foci 2x^2+y^2=3 I can't figure out how to put in (x^2/a^2)+(y^2/b^2)=1 form. How do I get rid of the 2 from 2x? and make 3 into 1?      Log On


   

Question 38654: ellipse problem #11 find vertices, foci
2x^2+y^2=3
I can't figure out how to put in (x^2/a^2)+(y^2/b^2)=1 form. How do I get rid of the 2 from 2x? and make 3 into 1?
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
2%28x%5E2%29+%2B+%28y%5E2%29+=+3
%282%28x%5E2%29%29%2F3+%2B+%28y%5E2%29%2F3+=+3%2F3
%28x%5E2%29%2F%283%2F2%29+%2B+%28y%5E2%29%2F3+=+1
Center is (0,0)
The major axis is vertical
a+=+sqrt%283%29
b+=+sqrt%281.5%29
c%5E2+=+a%5E2+-+b%5E2
c%5E2+=+%28sqrt%283%29%29%5E2+-+%28sqrt%281.5%29%29%5E2
c%5E2+=+%283%29+-+%281.5%29
c+=+sqrt%281.5%29
Vertices: (0,3),(0,-3),(1.5,0),(-1.5,0)
Foci: (0,1.5),(0,-1.5)