Noob having problems with standard form

Hi all :)

The below equation is a quadric. What is it in standard form?

5x^2 + 3y^2 + 3z^2 - 2xy + 2yz - 2xz - 10x + 6y - 2z - 9 = 0

Ok so in my book I understand the majority of how to do it. I get to simplifying the 'completing the square part' and this leaves

6(x'')^2 + 3(y'')^2 + 2(z'') - 18 = 0

This next part is the bit I do not understand what is going on.

The book says

"the equation of the quadric in standard form is:

(((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) = 1

this is the equation of an ellipsoid"

Could some explain the rule how it goes from

6(x'')^2 + 3(y'')^2 + 2(z'') - 18 = 0

to get

(((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) = 1

Re: Noob having problems with standard form

Hi kirstyaa9! :)

6(x'')^2 + 3(y'')^2 + 2(z'') - 18 = 0

Divide by 18:

(6/18)(x'')^2 + (3/18)(y'')^2 + (2/18)(z'') - (18/18) = 0

Rewrite:

(((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) - 1 = 0

(((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) = 1

Re: Noob having problems with standard form

Quote:

Originally Posted by

**ILikeSerena** Hi kirstyaa9! :)

6(x'')^2 + 3(y'')^2 + 2(z'') - 18 = 0

Divide by 18:

(6/18)(x'')^2 + (3/18)(y'')^2 + (2/18)(z'') - (18/18) = 0

Rewrite:

(((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) - 1 = 0

(((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) = 1

Thanks, didn't think it was that simple :).