Question 1204654
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I appreciate that you've shown your work so far.
You're on the right track. 
I'll start with the third step of your scratch work.


9(x^2 - x) + 25(y^2 - 2y) = 197.75
9(x^2 - x + <font color=red>0</font>) + 25(y^2 - 2y + <font color=red>0</font>) = 197.75
9(x^2 - x + <font color=red>0.25 - 0.25</font>) + 25(y^2 - 2y + <font color=red>1 - 1</font>) = 197.75
9((x^2 - x + 0.25) - 0.25) + 25((y^2 - 2y + 1) - 1) = 197.75
9((x-0.5)^2 - 0.25) + 25((y-1)^2 - 1) = 197.75
9(x-0.5)^2 + 9(-0.25) + 25(y-1)^2 + 25(-1) = 197.75
9(x-0.5)^2 - 2.25 + 25(y-1)^2 - 25 = 197.75
9(x-0.5)^2 + 25(y-1)^2 - 27.25 = 197.75
9(x-0.5)^2 + 25(y-1)^2 = 197.75 + 27.25
9(x-0.5)^2 + 25(y-1)^2 = 225


I'll let you finish up. The goal is to get the ellipse equation into the form {{{((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1}}}
(h,k) = center of the ellipse
a = half of the horizontal width
b = half of the vertical height
For example if a = 3 and b = 7 then the ellipse would be 2a = 2*3 = 6 units wide and 2b = 2*7 = 14 units tall.


In the steps where I've marked the terms in <font color=red>red</font>, I'm completing the square.
For the (x^2-x) portion, the x coefficient is -1. That cuts in half to -0.5 and squares to 0.25
We add and subtract 0.25 to help complete the square for the x terms.
The y terms are the same idea: cut the y coefficient (-2) in half to get -1, then it squares to 1.


Use a graphing tool like GeoGebra or Desmos to check your answer.



If a > b, then the foci are located at (h-c, k) and (h+c, k) where,
c^2 = a^2 - b^2
OR
If b > a, then the foci are located at (h, k-c) and (h, k+c) where,
c^2 = b^2 - a^2
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