SOLUTION: Graph the equation. Identify the vertices, co-vertices, and foci of the ellipse. 64x^2+16y^2=1024

Question 604754: Graph the equation. Identify the vertices, co-vertices, and foci of the ellipse. 64x^2+16y^2=1024
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Graph the equation. Identify the vertices, co-vertices, and foci of the ellipse. 64x^2+16y^2=1024
divide by 1024
x^2/16+y^2/64=1
This is an equation for an ellipse with vertical major axis.
Its standard form: (x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k)=(x,y) coordinates of center
..
For given equation: x^2/16+y^2/64=1
center:(0,0)
a^2=64
a=√64=8
vertices: (0,0�a)=(0,0�8)=(0,-8) and (0,8)
..
b^2=16
b=4
co-vertices: (0�b,0)=(0�4,0)=(0,-4) and (0,4)
..
c^2=a^2-b^2=64-16=48
c=√48≈6.9
foci: (0,0�c)=(0,0�6.9)=(0,-6.9) and (0,6.9)
...
see graph below:
y=�(64-4x^2)^.5

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