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

Question

the eccentricity of the ellipse 4x²+y²-8x+2y+4

Answer 1

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,y_1), 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

 

 

Answer 2

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