SOLUTION: Please help and show work. thanks Write the equation of the ellipse 36x^ 2 +9y^ 2 +360x−54y+657=0 in standard form (x−h)^ 2/ a^ 2 +(y−k)^ 2/ b^ 2 =1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please help and show work. thanks Write the equation of the ellipse 36x^ 2 +9y^ 2 +360x−54y+657=0 in standard form (x−h)^ 2/ a^ 2 +(y−k)^ 2/ b^ 2 =1      Log On


   

Question 999213: Please help and show work. thanks
Write the equation of the ellipse 36x^ 2 +9y^ 2 +360x−54y+657=0 in standard form (x−h)^ 2/ a^ 2 +(y−k)^ 2/ b^ 2 =1
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
to write the equation of the ellipse 36x%5E2+%2B9y%5E2+%2B360x-54y%2B657=0 in standard form %28x-h%29%5E2%2Fa%5E2+%2B%28y-k%29%5E2%2Fb%5E+2+=1, complete the squares, group all x and all y together

%2836x%5E2%2B360x%2Bb%5E2%29-b%5E2%2B%289y%5E2-54y%2Bb%5E2%29-b%5E2%2B657=0...factor out 36 from first group and 9 from second group

36%28x%5E2%2B10x%2Bb%5E2%29-36b%5E2%2B9%28y%5E2-6y%2Bb%5E2%29-9b%5E2%2B657=0....recall: %28a-b%29%5E2=a%5E2-2ab%2Bb%5E2
compare to 36%28x%5E2%2B10x%2Bb%5E2%29 and you see that a=1 and
2ab=10;solve for b
2%2A1%2Ab=10=>b=5
in %28y%5E2-6y%2Bb%5E2%29=>2ab=6=>b=3

so, we have

36%28x%5E2%2B10x%2B5%5E2%29-36%2A5%5E2%2B9%28y%5E2-6y%2B3%5E2%29-9%2A3%5E2%2B657=0

36%28x%2B5%29%5E2-36%2A25%2B9%28y-3%29%5E2-9%2A9%2B657=0

36%28x%2B5%29%5E2-900%2B9%28y-3%29%5E2-81%2B657=0

36%28x%2B5%29%5E2%2B9%28y-3%29%5E2-981%2B657=0

36%28x%2B5%29%5E2%2B9%28y-3%29%5E2-324=0

36%28x%2B5%29%5E2%2B9%28y-3%29%5E2=324....both sides divide by 324

36%28x%2B5%29%5E2%2F324%2B9%28y-3%29%5E2%2F324=324%2F324

36%28x%2B5%29%5E2%2F324%2B9%28y-3%29%5E2%2F324=1....simplify



%28x%2B5%29%5E2%2F9%2B%28y-3%29%5E2%2F36=1

=> h=-5, k=3, a=3, b=6