Help with Conics: Change General Form to Standard Form

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Homework Help Overview

The discussion revolves around converting the general form of a conic section to its standard form, specifically focusing on the equation 2x^2 + y^2 + 12x – 2y + 15=0. Participants express confusion regarding the process of factoring and completing the square.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various steps in completing the square and factoring, with some questioning the correctness of specific terms used in the process. There are attempts to clarify the need to factor out coefficients before completing the square.

Discussion Status

The conversation includes multiple interpretations of the steps needed to arrive at the standard form. Some participants provide guidance on how to correctly factor and complete the square, while others express uncertainty about their own calculations.

Contextual Notes

There is mention of confusion regarding the constants involved in the equation and the order of operations in factoring and completing the square. Participants are also navigating the constraints of homework rules that may limit their ability to seek direct answers.

ms. confused
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Ok I seem to be having problems with changing the general form of a conic to standard form. I'm mainly confused with how to factor, since I haven't done it in a while, as well as how to go about completing the square.

Here's one of my problems:

2x^2 + y^2 + 12x – 2y + 15=0

I rearranged it to look like: 2x^2 + 12x + y^2 – 2y + 15=0

Then I "supposedly" completed the square:

(2x^2 + 12x +36) + (y^2 – 2y +1)= 22

Factoring is where I got stuck: 2(x^2 + 6x +18) + (y-1)^2= 22

I don't know what to do with what I got and the answer is supposed to be:

(x+3)^2 / 2 + (y-1)^2 / 4 = 1
 
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(2x^2 + 12x +36)

Your problem is that this isn't a square! (Though, x^2 + 12x + 36 is) Your problem is you need to factor out the two first, so that the coefficient on x^2 is a 1.
 
I did that and I got 2(x^2 + 6x +18).
 
1. 2x^2 + 12x + 18 which is equivalent to 2(x+3)(x+3) + y^2 - 2y + 1 which is equivalnt to (y-1)(y-1) = -15 + 18 +1
2. Your equation is 2(x+3)^2 + (y-1)^2 = 4
3. Divide each side by 4. Now you have:
2(x+3)^2/4 + (y-1)^2/4 = 4/4
4. Now, your final product is:
(x+3)^2/2 + (y-1)^2/4 = 1

Is that the needed answer?
 
You need to factor before you figure out the constant term. You picked 36, then factored, which is the wrong way around.
 
How did you get 2x^2 + 12x + 18? I got 2x^2 + 12x + 36.
 
All you know is 2x^2 + 12x + ?.

Factor out the two to make it easy.

Now you have 2(x^2 +6x + ?)

Then you can fill in the square by making it 2(x^2 + 6x + 9) or 2(x+3)^2
 
This should give you 2x^2 + 12x + 18.

I hope I'm doing this right...
 
If you know the answer,then u can cheat:
1.Make in the initial quadratic form the 2 substitutions
x\rightarrow u-3
v\rightarrow v+1

2.Show that the new quadratic form is
\frac{u^{2}}{2}+\frac{v^{2}}{4}=1

3.Reverse the substitution and find the answer.

Daniel.
 
  • #10
Oh I see! Thanks for the help guys!
 

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