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Question 666318: What is the standard form of an ellipse with the equation x^2+9y^2-8x-90y+232=0?
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Given
(1) x^2 + 9y^2 -8x -90y +232 = 0
And you need to juggle it around to get it in the standard (useful) form of an ellipse given by
(2) ((x-h)^2))/a^2 + ((y-k)^2)/b^2 = 1
All you need to juggle is to use tha technique called "completing the square". Have you heard of it? Used it? It's very useful so watch the process as we solve your problem.
First reorganize (1) to put the x terms together then the y terms together, plus a constant equals zero as
(3) (x^2 - 8x) + (9y^2 - 90y) +232 = 0
Let's do the "complete the square" on the x terms first,
(4) x^2 -8x is the same as
(5) (x-8/2)^2 -16 or
(6) (x-4)^2 - 16, believe it? Let's check by simplifying (6).
Using FOIL on (6) yields
(7) x^2 -4x -4x +16 -16 becomes
(8) x^2 -8x, which is the same as (4).
Now complete the square on the y terms
(9) 9y^2 - 90y is
(10) 9*(y^2 -10y) which becomes
(11) 9 or
(12) 9*(y-5)^2 -9*25
Now place (6) and (12) into (3) and get
(13) (x-4)^2 -16 + 9*(y-5)^2 -9*25 + 232 = 0
Now collect all the constants in (13) to get
(14) (x-4)^2 + 9*(y-5)^2 = 16 +9*25 -232 or
(15) (x-4)^2 +9*(y-5)^2 = 9
Note that the standard form of the ellipse, given by (2), is equated to one, so we need to juggle one more time to get one on the right side of (15). Do this by simply dividing both sides of (15) by 9 and get
(16) ((x-4)^2)/9 + (y-5)^2 = 1
Comparing your answer of (16) to the standard form gives
(17) h=4, k=5, a=3, and b=1
The only way you can check your answer is to simplify (16) and see if it equals (1). Let's do it.
The first step I would do is get rid of the fraction 1/9, by multiplying both sides by 9. Then we need to simplify the parentheses and get
(18) (x^2 -8x +16) + 9*(y^2 -10y +25) = 9 or
(19) x^2 + 9y^2 -8x -90y +16 +225 -9 = 0 or
(20) x^2 + 9y^2 -8x -90y +232 = 0. Voila, we did it!
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