Help with Conics: Change General Form to Standard Form

  • Thread starter Thread starter ms. confused
  • Start date Start date
  • Tags Tags
    Conics
Click For Summary
SUMMARY

The discussion focuses on converting the general form of a conic section, specifically the equation 2x² + y² + 12x - 2y + 15 = 0, into its standard form. The key steps involve completing the square and factoring correctly. The final standard form is derived as (x+3)²/2 + (y-1)²/4 = 1. The critical takeaway is to factor out the leading coefficient before completing the square to simplify the process.

PREREQUISITES
  • Understanding of conic sections and their equations
  • Proficiency in completing the square method
  • Ability to factor quadratic expressions
  • Familiarity with algebraic manipulation techniques
NEXT STEPS
  • Learn how to complete the square for quadratic equations
  • Study the properties and graphs of conic sections
  • Explore advanced factoring techniques for polynomials
  • Practice converting between general and standard forms of conic equations
USEFUL FOR

Students studying algebra, educators teaching conic sections, and anyone looking to improve their skills in algebraic manipulation and quadratic equations.

ms. confused
Messages
90
Reaction score
0
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
 
Physics news on Phys.org
(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!
 

Similar threads

Replies
27
Views
1K
  • · Replies 40 ·
2
Replies
40
Views
2K
Replies
3
Views
987
Replies
10
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
24
Views
4K
Replies
13
Views
1K