SOLUTION: Find the center, vertices, foci, and the lengths of the major and minor axes of the ellipse. {{{4x^2+9y^2=36y}}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the center, vertices, foci, and the lengths of the major and minor axes of the ellipse. {{{4x^2+9y^2=36y}}}      Log On


   

Question 813151: Find the center, vertices, foci, and the lengths of the major and minor axes of the ellipse.
4x%5E2%2B9y%5E2=36y
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the center, vertices, foci, and the lengths of the major and minor axes of the ellipse.4x%5E2%2B9y%5E2=36y
***
4x^2+9y^2=36y
4x^2+9y^2-36y=0
complete the square:
4x^2+9(y^2-4y+4)=0+36
4x^2+9(y-2)^2=36
divide by 36
x^2/9+(y-2)^2/4=1
Given ellipse has a horizontal major axis.
Its standard form of equation:%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1, a>b, (h,k)=(x,y) coordinates of center
..
center: (0,2)
a^2=9
a=3
length of horizontal major axis=2a=6
vertices: (0±a,2)=(0±3,2)=(-3,2) and (3,2)
..
b^2=4
b=2
length of minor axis=2b=4
..
c^2=a^2-b^2=9-4=5
c=√5≈2.2
foci:(0±c,2)=(0±2.2,2)=(-2.2,2) and (2.2,2)