Question 541736: for the equation of the given ellipse 16x^2+4y^2=64. I need help finding the verticies, x or y-intercepts, foci, and how to sketch this also.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! for the equation of the given ellipse 16x^2+4y^2=64. I need help finding the verticies, x or y-intercepts, foci, and how to sketch this also.
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16x^2+4y^2=64
divide by 64
x^2/4+y^2/16=1
This is an equation for an ellipse with vertical major axis of the standard form:
(x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k)=(x,y) coordinates of the center
..
For given ellipse:
center: (0,0)
a^2=16
a=4
Vertices: (0, 0±a)=(0,0±4)=(0,-4) and (0,4)
..
b^2=4
b=2
..
c^2=a^2-b^2=16-4=12
c=√12
Foci: (0, 0±c)=(0,0±√12)=(0,-√12) and (0,√12)
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x-intercepts
set y=0, then solve for x
16x^2+4y^2=64
16x^2=64
x^2=4
x=±2
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y-intercepts
set x=0, then solve for y
16x^2+4y^2=64
4y^2=64
y^2=16
y=±4
..
See graph below as a visual check on answers above:
y=±(16-4x^2)^.5
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