Question 744237: How do i find the vertices and foci of; (x^2/64)-(9y^2/4)=1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! How do i find the vertices and foci of; (x^2/64)-(9y^2/4)=1
Given ellipse has a horizontal major axis.
Its standard form of equation:  ,a>b, (h,k)=(x,y) coordinates of center
(x^2/64)-(9y^2/4)=1
(x^2/64)-y^2/(4/9)=1
center: (0,0)
a^2=64
a=8
vertices: (0±a,0)=(0±8,0)=(-8,0) and (8,0)
b^2=4/9
b=2/3
c^2=a^2-b^2=64-4/9=(576/9)-(4/9)=572/9
c=√(572/9)≈7.97
foci: (0±c,0)=(0±7.97,0)=(-7.97,0) and (7.97,0)
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