>>...this is an Ellipses problem but I can't figure it out. 
Could you please help me. x^2+4y^2-10x-40y+121=0...<<
          x² + 4y² - 10x - 40y + 121 = 0
Group x terms together and y-terms together and get the 121 off the left side:
                x² - 10x + 4y² - 40y = -121 
Factor the coefficients of x and y out of the pairs of terms in each 
variable:
           1(x² - 10x) + 4(y² - 10y) = -121
Complete the square in each of the parentheses.  Take half the 
coefficients of x or y, respectively, and square them.  We need to 
add 25 in the first parentheses and 25 in the second.  To offset this 
we must add 1∙25 and 4∙25 to the right side:
 1(x² - 10x + 25) + 4(y² - 10y + 25) = -121 + 1∙25 + 4∙25  
Factor the trinomials in the parentheses as perfect squares:
               1(x - 5)² + 4(y - 5)² = -121 + 25 + 100
                (x - 5)² + 4(y - 5)² = 4
Divide every term through by 4 to get 1 on the right
              
              

a² > b², so this is of the form 

              

an ellipse with horizontal major axis,
which has center at (h,k) = (5,5), a²=4, so a=2 = semi-major axis,
b²=1, so b=1 = semi-minor axis.   


Edwin