# The equation of a hyperbolic paraboloid to derive the corner points of rectangle

#### bugatti79

##### Member
Hi Folks,

I have come across some text where $$f(x,y)=c_1+c_2x+c_3y+c_4xy$$ is used to define the corner points

$$f_1=f(0,0)=c_1$$
$$f_2=f(a,0)=c_1+c_2a$$
$$f_3=f(a,b)=c_1+c_2a+c_3b+c_4ab$$
$$f_4=f(0,b)=c_1+c_3b$$

How are these equations determined? F_1 to F_4 starts at bottom left hand corner and rotates counter clockwise.

Thanks
Note: This was posted last week on MHB with no response.
The equation of a hyperbolic paraboloid to derive the corner points of rectangle

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hi Folks,

I have come across some text where $$f(x,y)=c_1+c_2x+c_3y+c_4xy$$ is used to define the corner points

$$f_1=f(0,0)=c_1$$
$$f_2=f(a,0)=c_1+c_2a$$
$$f_3=f(a,b)=c_1+c_2a+c_3b+c_4ab$$
$$f_4=f(0,b)=c_1+c_3b$$

How are these equations determined? F_1 to F_4 starts at bottom left hand corner and rotates counter clockwise.

Thanks
Note: This was posted last week on MHB with no response.
The equation of a hyperbolic paraboloid to derive the corner points of rectangle
Hi bugatti79! I think you meant MHF instead of MHB. ;-)

Your f(x,y) is defined on a square domain of the form [0,a] x [0,b].
The corner points are the corner points of this domain.

#### bugatti79

##### Member
Hi bugatti79! I think you meant MHF instead of MHB. ;-)

Your f(x,y) is defined on a square domain of the form [0,a] x [0,b].
The corner points are the corner points of this domain.
Hello Sir,

Yes I stand corrected regarding MHF. Ok but just wondering how those equations were derived particularly the third one with the xy product term #### Ackbach

##### Indicium Physicus
Staff member
If you take a general hyperbolic paraboloid and rotate it appropriately, you can get its form to look like your $f(x,y)$. It's entirely analogous to being able to take a certain hyperbola, rotate it, and obtain the equation $xy=1$. I haven't worked it out, but there probably is a standard-form hyperbolic paraboloid equation looking something like
$$\frac{z}{c}= \frac{y^{2}}{b^{2}}- \frac{x^{2}}{a^{2}},$$
which, when rotated using rotation matrices, gives you the form of your $f(x,y)$.

#### bugatti79

##### Member
If you take a general hyperbolic paraboloid and rotate it appropriately, you can get its form to look like your $f(x,y)$. It's entirely analogous to being able to take a certain hyperbola, rotate it, and obtain the equation $xy=1$. I haven't worked it out, but there probably is a standard-form hyperbolic paraboloid equation looking something like
$$\frac{z}{c}= \frac{y^{2}}{b^{2}}- \frac{x^{2}}{a^{2}},$$
which, when rotated using rotation matrices, gives you the form of your $f(x,y)$.
Hi Ackbach,
Good to see you are still around! That is interesting. I can see now why this expression is required. Each corner needs to movable while the other three are fixed thus resembling the shape of hyperbolic parabola from a flat surface.

I googled information on rotation of a hyperbolic parabola to a flat surface and vica versa, no luck. Perhaps some one know of a link?

Thanks guys

#### Ackbach

##### Indicium Physicus
Staff member
Good to see you around, too! I remember you having the most interesting problems in the old country - quantum mechanics and such-like.

Well, I just got my info from the wiki on hyperbolic paraboloids. There's the general equation right away, and then a little info on rotation in the "Multiplication Table" section. To be clear: you're not rotating the hyperbolic paraboloid into the flat surface. A hyperbolic paraboloid is not flat in a Euclidean sense! If I had to make a guess, I'd say the rectangle you're dealing with is some region of interest for other reasons.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Not to be confused with quartics or quadratics.

In your case you have a hyperbolic paraboloid that has been rotated around the z-axis by 45 degrees.

#### bugatti79

##### Member
Not to be confused with quartics or quadratics.

In your case you have a hyperbolic paraboloid that has been rotated around the z-axis by 45 degrees.
Hi,

Thanks for that. Where did you establish that regarding the rotation of 45degrees?

Ed

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hi,

Thanks for that. Where did you establish that regarding the rotation of 45degrees?

Ed
I'll give a summarized analysis.

Your quadric equation is $c_1 + c_2x + c_3y + c_4xy - z = 0$.
This can be written as the following quadric equation in matrix form (see quadric on wiki):
$$\begin{bmatrix}x & y & z\end{bmatrix} \begin{bmatrix}0 & c_4/2 & 0 \\ c_4/2 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} + \begin{bmatrix}c_2 \\ c_3 \\ -1\end{bmatrix} \cdot \begin{bmatrix}x \\ y \\ z\end{bmatrix} = 0$$

The eigenvalues of the matrix are $\dfrac {c_4} 2, -\dfrac {c_4} 2, 0$ and their corresponding eigenvectors are $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$.

Those eigenvectors are diagonals, showing a rotation of 45 degrees.

$$\frac {c_4}{2} x^2 - \frac {c_4}{2}y^2 - C z = 0$$
where the constant C is yet to be determined (but it's 1).

As you can see in the article, this is classified as a hyperbolic paraboloid.

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#### bugatti79

##### Member
I'll give a summarized analysis.

Your quadric equation is $c_1 + c_2x + c_3y + c_4xy - z = 0$.
This can be written as the following quadric equation in matrix form (see quadric on wiki):
$$\begin{bmatrix}x & y & z\end{bmatrix} \begin{bmatrix}0 & c_4/2 & 0 \\ c_4/2 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} + \begin{bmatrix}c_2 \\ c_3 \\ -1\end{bmatrix} \cdot \begin{bmatrix}x \\ y \\ z\end{bmatrix} = 0$$

The eigenvalues of the matrix are $\dfrac {c_4} 2, -\dfrac {c_4} 2, 0$ and their corresponding eigenvectors are $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$.

Those eigenvectors are diagonals, showing a rotation of 45 degrees.

$$\frac {c_4}{2} x^2 - \frac {c_4}{2}y^2 - C z = 0$$
So far how I understand it is that we start off with the quadric surface $$f(x,y)=c_1+c_2x+c_3y+c_4xy$$, rotate it 45 degrees as Serena showed which gives us the non degenerate form. If we set the A abd B coefficients to 0 then z=0 and thus the surface becomes flat (Use applet)...