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March 4, 2015

March 4, 2015

Posted by **Emily** on Monday, November 23, 2009 at 11:21pm.

34. x^2 + 25y^2 - 6x - 100y + 84 = 0

(Hint: Complete the squares in x and y. Begin by writing the equation in this form: (x^2 - 6x + ___) + 25(y^2 - 4y + ___) = -84.)"

==> I used the hint that came with the problem, but as I tried to solve the equation into the general form for an ellipse, I was getting negative denominators, which would make a and b contain imaginary numbers. Does anyone know how to solve this? Any help is greatly appreciated! :)

- Precalculus -
**Reiny**, Monday, November 23, 2009 at 11:38pmx^2 + 25y^2 - 6x - 100y + 84 = 0

x^ - 6x + 9 + 25(y^2 - 4y + 4) = -84 + 9 + 100

notice in the y expression, I added 4 to the inside, but that was multiplied by 25, so I had to add 100 on the right side.

(x-3)^2 + 25(y-2)^2 = 25

(x-3)^2/25 + (y-2)^2 = 1

so a^2 = 25, and b^2 = 1

take it from there

- Precalculus -
**Emily**, Monday, November 23, 2009 at 11:59pmOhhhhh!!!!! This totally makes sense. I forgot to multiply the 4 by 25 before bringing it over to the other side. Thanks so much!! :D

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