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Question 1204654: For the ellipse, determine the
a) coordinates of the center
b) lengths of the major and minor axes
c) coordinates of the foci
9x^2 + 25y^2 - 9x - 50y - 197.75 = 0
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I did not get very far in this question. I was trying to first complete the square to factor into the standard form of the elliptical equation and move on from there.
9x^2 + 25y^2 - 9x - 50y - 197.75 = 0
9x^2 - 9x + 25y^2 - 50y = 197.75
9(x^2 - x) + 25(y^2 - 2y) = 197.75
9(x^2 - x + (1/4)) + 25(y^2 - 2y +1) = 197.75 + (1/4) + 1
9(x - (1/2))^2 + 25(y - 1)^2 = 199
I don’t know where to go from here because the numbers are troubling me
Found 3 solutions by greenestamps, math_tutor2020, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
What's this?! A reader who actually showed in his post the work he has done on the problem??!! Amazing....
Thank you!
I see you know what you are doing, so I will just point out your error and let you continue working on the problem.
When you complete the square, you show, on the right side, adding (1/4) + 1.
What you should be adding is 9(1/4) + 25(1). That's a very common error, even with people experienced with the process of completing the square, if they aren't paying attention....
That will make the number on the right 225, which is 15^2, making it easy to finish the problem with "nice" numbers.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
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 + 0) + 25(y^2 - 2y + 0) = 197.75
9(x^2 - x + 0.25 - 0.25) + 25(y^2 - 2y + 1 - 1) = 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 
(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 red, 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
Answer by ikleyn(52795)  (Show Source):
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