Conics. Finding standard form of non-origin centered equa...

A

~AD~

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Hello, I am new here and I just cannot understand how to do this :(

I am looking at the examples in my book, and I do not understand how they got from point A to B. I need to know this because there are questions like this on my assignments that are due soon. For example:

\(\displaystyle y^2 - 4x^2 + 4y + 24x - 41 = 0\)

and the answerr in standard for would be:

\(\displaystyle [(y + 2)^2 / 3^2}] - [(x - 3)^2 / (3/2)^2] = 1\)

I am reading here that it says to 'complete the square to write equation in standard form.' I managed to stay with them all the way to step 4. Which i do not understand where they get these numbers...
Assuming the original equation is step 1, heres the rest:

2) Group terms (y^2 + 4y + __ ) - (4x^2 - 24x + __ ) = 41 // Ok, I'm with them so far...

3) Factor 4 out of x-terms (y^2 + 4y + __ ) - 4(x^2 - 6x + __ ) // Ok... are you allowed to factor out of y-terms also if you could??

4) Add 4 and Subtract 4(9) (y^2 + 4y + 4) - 4(x^2 - 6x + 9) = 41 + 4 - 4(9) // Ok???? Where did they get these numbers from???



And from that point I just lost it... I get similar stuff doing Ellipses and Circles....The above equation is a Hyperbola, by the way.


Gladly accept any help... thanks...
 
~AD~ said:
3) Factor 4 out of x-terms (y^2 + 4y + __ ) - 4(x^2 - 6x + __ ) // Ok... are you allowed to factor out of y-terms also if you could??
Click to expand...


You could, but you don't want to in this case because the coefficient on y^2 is already 1.

(The first step in the process called Completing the Square is to make sure that the coefficient on the squared term is 1.)

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4) Add 4 and Subtract 4(9) (y^2 + 4y + 4) - 4(x^2 - 6x + 9) = 41 + 4 - 4(9) // Ok???? Where did they get these numbers from???
Click to expand...


The numbers added on the left-hand side of the equation is where the actual squares are completed. You add the specific number necessary to obtain a perfect square trinomial. Here's how you determine the number:

Given x^2 + bx

Add (b/2)^2

So, you have x^2 + bx + (b/2)^2

This always factors as (x + b/2)^2



EG:

Given x^2 + 6x

Add (6/2)^2

So, you have x^2 + 6x + 9

This factors as (x + 3)^2



The numbers on the right-hand side are added to keep the equation balanced, of course.

It's unfortunate that nobody made sure you understood completing the square before trying to teach you to work with conic sections because it's an easy process to understand (vital, too). :(

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Click here to see several of thousands of Internet lessons.
 
"completing the squares" is a common method of solving quadratic equation - which is used for other purposes too (like for this problem).

For an excellent review, go to:

http://www.purplemath.com/modules/sqrquad.htm

and then do further google search with the keywords (completing the squares).
 
Thanks guys!! :D Makes sense now. And I think they did teach us completing the square during our quadratics lessons. But he said we can use many ways to solve, and completing the square wasn't a favorite. So I never used it. I never figured I'd need it again. :p :oops: Thanks again!
 
A word frm somebody who has suffered through "Never-will-use-it-in-my-life" syndrome ---

Don't say it - don't think it!!!

Exactly three semester after you have finished thinking that - BAM!! - you'll be hit with that.

Remember - the amount that you need to learn these days to catch up with the world is enormous. Teachers really don't like to waste time on teaching you useless things.
 
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