SOLUTION: 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

>>...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