Proof of Quadratic Surfaces of 2 Variables

[dumkid]

Please I really need help with this homework... I need to prove this theorem.

Consider the quadratic function: f(x,y) = Ax^2 + 2Bxy + Cy^2. Introduce the following quantity: D = AC - B^2 (The Discriminant) There are only 3 possibilities for the graph f(x,y)

1. If D = AC-B^2 > 0 then the graph is elliptical paraboloid
- If A > 0 (And C > 0) then opens upward
- If A < 0 (And C < 0) then opens downward
2. If D = AC-B^2 < 0 then the graph is a saddle
3. If D = AC-B^2 = 0 then the graph is a cylinder

I need to prove this... I do not know where or how to start or how to finish.. It is really hard for me to understand the assignment. I am really bad at proving theorems. The actual assignment can be viewed here:

http://img525.imageshack.us/img525/9722/18774399hr2.jpg < Pg1
http://img396.imageshack.us/img396/9646/85916334yz5.jpg < Pg2
http://img183.imageshack.us/img183/3959/64328990ku0.jpg < Pg3

I know I am given a lot of hints... but maybe I am making it more complicated than it actually is. It does not make sense to me... Ive read it.. I just need guidance of how I should begin... and how it should end. Maybe if anyone knows where this was proven before... can give me a link that would be great too... I've searched all of google... nothing in book about it... Thank you!

 

[Deleted member 4993]

Consider the quadratic function: f(x,y) = Ax^2 + 2Bxy + Cy^2. Introduce the following quantity: D = AC - B^2 (The Discriminant). There are only 3 possibilities for the graph f(x,y)

1. If D = AC-B^2 > 0 then the graph is elliptical paraboloid
- If A > 0 (And C > 0) then opens upward
- If A < 0 (And C < 0) then opens downward
2. If D = AC-B^2 < 0 then the graph is a saddle
3. If D = AC-B^2 = 0 then the graph is a cylinder !

Can you prove similar problem in 2_D. That is.

Consider the quadratic function: f(x) = Ax^2 + 2Bx + C. Introduce the following quantity: D = AC - B^2 (The Discriminant)
There are only 3 possibilities for the graph f(x):

1. If D = AC-B^2 > 0 then the graph does not intersect x-axis
- If A > 0 (And C > 0) then opens upward
- If A < 0 (And C < 0) then opens downward
2. If D = AC-B^2 < 0 then the graph intersects x-axis
3. If D = AC-B^2 = 0 then the graph is tangent to x-axis

Now think - what does the introduction of the third dimension do to the graph?

 

[dumkid]

It will pull/strech the graph upward?

But how do the 4 statements (pg 2-3 on assignment) help prove the theorem? how would I do it?

Thanks

 

[Deleted member 4993]

It will pull/strech the graph upward?

Think what type of graph will be there at x = constant or y = constant

Attack #3 first

\(\displaystyle B^2 \, = \, A\cdot C\)

\(\displaystyle B \, = \, \sqrt{A\cdot C}\)

\(\displaystyle f(x,y) \, = \, A\cdot x^2 \, + \, 2\cdot B\cdot x\cdot y \, + \, C\cdot y^2 \, = \, A\cdot x^2 \, + \, 2\cdot \sqrt{A\cdot C}\cdot x\cdot y \, + \, C\cdot y^2\)

Now continue................

 

[dumkid]

Is that the equation of the line for 3) D=0 ?

I still dont understand... anyone else?

 

[Deleted member 4993]

You need to show me - what you have done so far.

Start with definition elliptic paraboloid - what is it?

For a parabola - how did you prove it was opening up or down?