Is there a transformation if at all?

[flyingpig]

Homework Statement

Consider

Ax² + Bxy + Cy² + Dx + Ey = F

What happens if

Ax² + B(x + h)(y + k) + Cy² + Dx + Ey = F

The Attempt at a Solution

Does it do anything?

 

[Mark44]

Homework Statement

Consider

Ax² + Bxy + Cy² + Dx + Ey = F

What happens if

Ax² + B(x + h)(y + k) + Cy² + Dx + Ey = F

The Attempt at a Solution

Does it do anything?

I'll say yes, but I wouldn't say this is any kind of normal transformation, since you are replacing only some x and y values by x + h and y + k.

 

[vela]

Multiply it out and write it in the form

A'x² + B'xy + C'y² + D'x + E'y = F'

That should shed some light on the effect of modifying the cross term.

 

[SammyS]

Homework Statement

Consider

Ax² + Bxy + Cy² + Dx + Ey = F

What happens if

Ax² + B(x + h)(y + k) + Cy² + Dx + Ey = F

The Attempt at a Solution

Does it do anything?

If what you say is taken literally, it means that

xy = (x + h)(y + k) .

 

[flyingpig]

@Sammy

Only if h and k are 0.

@vela, I would but I am moving back to the college and i packed everything and it looked a little tedious. I will get back to that diff and continu thread when I got my stuff set.

@Mark, please come back to my resume thread lol

Also this is just a "theory" before application. I am just going to guess that B(h + k) will get absorbed into the F

So that B(h + k) + F = K?
And so do the other terms get abosrbed.

 

[SammyS]

@Sammy

Only if h and k are 0.

So, that means that A, B, C, D, E, F do not represent the same set of numbers in the first equation compared to the second equation.

So, you should use primes or subscripts or some such scheme to differentiate (not in the Calculus sense) the two cases.