Question 511723: please help me with this, find the vertices, foci and length of major and minor axis :
x2+2y2-6x-8y=1
I know how to find the vertices, foci and length of major and minor but I have no idea how to complete the square for the eq, can you please show me.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find the vertices, foci and length of major and minor axis :
x2+2y2-6x-8y=1
**
x^2+2y^2-6x-8y=1
complete the square
(x^2-6x+9)+2(y^2-4y+4)=1+9+8=18
(x-3)^2+2(y-2)^2=18
divide by 18
(x-3)^2/18+(y-2)^2/9=1
This is an equation of an ellipse with horizontal major axis of the standard form:
(x-h)^2/a^2+(y-k)^2/b^2=1, a>b, (h,k) being the (x,y) coordinates of the center
For given equation: (x-3)^2/18+(y-2)^2/9=1
center: (3,2)
a^2=18
a=√18=4.24
length of major axis=2a=2√18=8.49
vertices=(3±a,2)=(3±√18,2)=(3+√18,2) and (3-√18,2)
..
b^2=9
b=√9=3
length of minor axis=2b=6
..
Foci
c^2=a^2-b^2=18-9=9
c=√9=3
Foci=(3±c,2)=(3±3,2)=(6,2) and (0,2)
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