Question 759821
x^2 + 4y^2 - 4x + 8y + 4 = 0
x^2 - 4x + 4y^2 + 8y +4 = 
(x^2 - 4x + 4) + 4(y^2 + 2y + 1) = 0
(x-2)^2 - 4 + 4(y + 1)^2 = 0
(x - 2)^2 + 4(y + 1)^2 = 4
<br>
divided by 4 so we get the right side become 1 according to ellipse equation:  
(x - 2)^2/4 + (y + 1)^2 = 1
<br>
The equation also can be written by:   
(x - 2)^2/4 + (y + 1)^2/1 = 1
<br>
we get:
<br>
h=2
k=-1
a^2=4 -> a=2
b^2=1 -> b=1
<br>
center of ellipse:
(h,k)
(2,-1)
<br>
vertices of ellipse:
(aħh,k)
(2ħ2,-1)
(2+2,-1) and (2-2,-1)
(4,-1) and (0,-1)