SOLUTION: Find the center, foci, vertices, and lengths of the major and minor axes.
16x^2 - 64x + 4y^2 = 0
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Question 885229: Find the center, foci, vertices, and lengths of the major and minor axes.
16x^2 - 64x + 4y^2 = 0
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find the center, foci, vertices, and lengths of the major and minor axes.
16x^2 - 64x + 4y^2 = 0
***
16x^2-64x+4y^2=0
complete the square:
16(x^2-4x+4)+4y^2=0+64
16(x-2)^2+4y^2=64
This is an equation of an ellipse with vertical major axis.
Its standard form of equation: (x-h)^2/a^2+(y-k)^2=1,
center:(2,0)
vertices:
a^2=16
a=4
length of major axis=2a=8
vertices:(2,0±a),(2,0±4),(2,-4)and (2,4)
b^2=4
b=2
length of minor axis=2b=4
c^2=a^2-b^2=16-4=12
c=√12≈3.5
foci:(2,0±c),(2,0±3.5),(2,-3.5)and (2,3.5)
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