Question 767746: 4x^2+y^2-48x-4y+48=0 find center C, length of major axis, length of minor axis, distance from C to foci?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 4x^2+y^2-48x-4y+48=0 find center C, length of major axis, length of minor axis, distance from C to foci?
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4x^2+y^2-48x-4y+48=0
complete the square:
4x^2-48x+y^2-4y+48=0
4(x^2-12x+36)+(y^2-4y+4)=-48+144+4
4(x-6)^2+(y-2)^2=100
(x-6)^2/25+(y-2)^2/100=1
This is an equation of an ellipse with vertical major axis
Its standard form:  , a>b, (h,k)=(x,y) coordinates of the center
For given ellipse:
center: (6,2)
a^2=100
a=√100=10
length of major axis=2a=20
b^2=25
b=√25=5
length of minor axis=2b=10
c^2=a^2-b^2=100-25=75
c=√75
distance from center to foci=√75≈8.66
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