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the ecentricity of the ellipse 4x^2+y^2-8x+2y+4

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2 Answers

To your expression an equation, describing the ellipse, you have to put
 
                                             4x2 +y2 - 8x +2y +4 = 0                                (1)
 
We can rewrite the expression in the following way:
 
                        4x2 - 8x + 4 +y2 +2y +1 -1 = 0       or
 
                           4(x-1)2 + (y+1)2 = 1                                                       (2)
 
If we write the equation for an ellipse in standard way
 
                            ( x-x1)2/a2   + (y-y1)2/b2    = 1                                      (3)
 
we can see that the center of the ellipse is at the point (x1,y1), x1 = 1, y1 = -1   and semi-axis are
a =1/2  and b =1, which means this ellipse is stretched along its vertical axis. The eccentricity of the
ellipse can be calculated by the formula
 
 
                             e =(√(b2 - a2) )/b    = √3/2 = 0.866
 
 
4x2 + y2 - 8x + 2y + 4 = 0
 
4x2 - 8x + 4 + y2 + 2y + 1 = 1
 
4(x - 1)2 + (y+1)2 = 1
 
We have: a = 1/2 (semiminor axis), b=1 (semimajor axis)
 
Compute c = √(b2 - a2) = (1/2)√3
 
Finally, ε = c/b = (1/2)√3 ≈ 0.8660254037844386