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
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
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