# How is z=2xy a Hyperbolic Paraboloid in the rotated 45° in the xy-plane?

• MHB
• WMDhamnekar
In summary: In the case of $z=2xy$ we have $B=\begin{bmatrix}1/\sqrt{2}&-1/\sqrt{2}&0\\1/\sqrt{2}&1/\sqrt{2}&0\\0&0&1\end{bmatrix}$ with determinant +1 as intended.So:$$\operatorname{Tr}(B)=1/\sqrt{2}+1/\sqrt{2}+1=1+2\cos\phi\implies \phi=\pm \frac\pi 4$$In summary, quadric surfaces can be translated and/or rotated so that their equations match one of the six types of quadric surfaces
WMDhamnekar
MHB
How to prove that every quadric surface can be translated and/or rotated so that its equation matches one of the six types of quadric surfaces namely 1) Ellipsoid

2)Hyperboloid of one sheet

3) Hyperboloid of two sheet 4)Elliptic Paraboloid

5) Elliptic Cone 6) Hyperbolic Paraboloid

The equation of hyperbolic parabolid is given by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=\frac{z}{c}\tag{1}$

Now, for example, z=2*x*y is a case of equation of mixed variables of the form $Ax^2+By^2+Cz^2+Dxy+ Exz +Fyz + Gx + Hy +Iz + J =0$ with e.g $D\not=0$ so that we get xy term. This equation does not match any of the equations of quadric surfaces, we know.However, by rotating the x- and y- axes by $45^\circ$ in the xy-plane by means of coordinate transformation $x=\frac{(x'-y')}{\sqrt{2}},y=\frac{(x'+y')}{\sqrt{2}}, z=z'$ then $z=2xy$ becomes hyperbolic paraboloid $z'=(x')^2-(y')^2$ in the (x',y',z') coordinate system.

That is z=2xy is a hyperbolic paraboloid as in equation (1) but rotated $45^\circ$ in the xy-plane.How to prove $$z=2*x*y$$ is a hyperbolic paraboloid?

Last edited:
Re: How $$z=2*x*y$$ is a Hyperbolic Paraboloid in the rotated $$45^\circ$$ in the xy-plane?

Dhamnekar Winod said:
How to prove $$z=2*x*y$$ is a hyperbolic paraboloid?

Rewrite as:
$$z=\begin{pmatrix}x&y&z\end{pmatrix}\begin{bmatrix}&1&\\1\\&&0\end{bmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$$

Diagonalize the matrix to find $\begin{bmatrix}1&&\\&-1\\&&0\end{bmatrix}$.
It means we can write the equation as a rotation (by 45 degrees in this case), which then results in the transformed equation
$$Z=X^2-Y^2$$
which is a hyperbolic paraboloid.

Re: How $$z=2*x*y$$ is a Hyperbolic Paraboloid in the rotated $$45^\circ$$ in the xy-plane?

Klaas van Aarsen said:
Rewrite as:
$$z=\begin{pmatrix}x&y&z\end{pmatrix}\begin{bmatrix}&1&\\1\\&&0\end{bmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$$

Diagonalize the matrix to find $\begin{bmatrix}1&&\\&-1\\&&0\end{bmatrix}$.
It means we can write the equation as a rotation (by 45 degrees in this case), which then results in the transformed equation
$$Z=X^2-Y^2$$
which is a hyperbolic paraboloid.
Hello,

One more question relating to this thread.

Is there any general formula establishing the relationship between p=(x,y,z) in the old coordinates system and p'= (x',y',z') in the new coordinates system if we transform the old coordinates system into new coordinates system by any degrees(radians) of angle? Say between 0 to 360 degrees or $2\pi$ radians

Last edited:
Re: How $$z=2*x*y$$ is a Hyperbolic Paraboloid in the rotated $$45^\circ$$ in the xy-plane?

Dhamnekar Winod said:
Hello,

One more question relating to this thread.

Is there any general formula establishing the relationship between p=(x,y,z) in the old coordinates system and p'= (x',y',z') in the new coordinates system if we transform the old coordinates system into new coordinates system by any degrees(radians) of angle? Say between 0 to 360 degrees or $2\pi$ radians

Yes.
Write the equation as $\mathbf x^T Q \mathbf x + P\mathbf x + R = 0$ where $Q$ is a symmetric matrix.

Then $Q$ is diagonalizable and there is an orthonormal basis of eigenvectors.
It means that we can write the matrix as $Q=BDB^T$, where $D$ is a diagonal matrix and $B$ is an orthogonal matrix.

If $Q$ is invertible, as it is for ellipsoids and hyperboloids, we can find the translation vector $\mathbf t$ with:
$$\mathbf t = -\frac 12 Q^{-1} P^T$$

We now have:
$$\mathbf x' = B^T(\mathbf x - \mathbf t)$$
as the relationship between the new coordinates and the old coordinates.

The matrix $B$ is the basis transformation matrix.
In 2D or 3D we can pick it to be a rotation matrix. (We can easily convert it to one if it's not already.)
Assuming it's a rotation matrix (determinant +1), we can then find the rotation angle $\phi$ by taking the trace ($\operatorname{Tr}$) of $B$, which is the sum of the elements on its diagonal.
If we ensure that $B$ has determinant +1, then in 2D we have $\operatorname{Tr}(B)=2\cos\phi$, and in 3D we have $\operatorname{Tr}(B)=1+2\cos\phi$.
Solve to find the angle $\phi$, which is determined up to a sign now.

In the case of $z=2xy$ we have $B=\begin{bmatrix}1/\sqrt{2}&-1/\sqrt{2}&0\\1/\sqrt{2}&1/\sqrt{2}&0\\0&0&1\end{bmatrix}$ with determinant +1 as intended.
So:
$$\operatorname{Tr}(B)=1/\sqrt{2}+1/\sqrt{2}+1=1+2\cos\phi\implies \phi=\pm \frac\pi 4$$

Re: How $$z=2*x*y$$ is a Hyperbolic Paraboloid in the rotated $$45^\circ$$ in the xy-plane?

Klaas van Aarsen said:
Yes.
Write the equation as $\mathbf x^T Q \mathbf x + P\mathbf x + R = 0$ where $Q$ is a symmetric matrix.

Then $Q$ is diagonalizable and there is an orthonormal basis of eigenvectors.
It means that we can write the matrix as $Q=BDB^T$, where $D$ is a diagonal matrix and $B$ is an orthogonal matrix.

If $Q$ is invertible, as it is for ellipsoids and hyperboloids, we can find the translation vector $\mathbf t$ with:
$$\mathbf t = -\frac 12 Q^{-1} P^T$$

We now have:
$$\mathbf x' = B^T(\mathbf x - \mathbf t)$$
as the relationship between the new coordinates and the old coordinates.

The matrix $B$ is the basis transformation matrix.
In 2D or 3D we can pick it to be a rotation matrix. (We can easily convert it to one if it's not already.)
Assuming it's a rotation matrix (determinant +1), we can then find the rotation angle $\phi$ by taking the trace ($\operatorname{Tr}$) of $B$, which is the sum of the elements on its diagonal.
If we ensure that $B$ has determinant +1, then in 2D we have $\operatorname{Tr}(B)=2\cos\phi$, and in 3D we have $\operatorname{Tr}(B)=1+2\cos\phi$.
Solve to find the angle $\phi$, which is determined up to a sign now.

In the case of $z=2xy$ we have $B=\begin{bmatrix}1/\sqrt{2}&-1/\sqrt{2}&0\\1/\sqrt{2}&1/\sqrt{2}&0\\0&0&1\end{bmatrix}$ with determinant +1 as intended.
So:
$$\operatorname{Tr}(B)=1/\sqrt{2}+1/\sqrt{2}+1=1+2\cos\phi\implies \phi=\pm \frac\pi 4$$

Hello,What is translation vector t ? and what is x' ?

In the case of z=2*x*y, what is translation vector t and what is x' ? If x' is a new coordinates system, then how is it equal to $B^T(x-t)?$My other queries(if any) will be posted in my next reply.:)

Last edited:
Re: How $$z=2*x*y$$ is a Hyperbolic Paraboloid in the rotated $$45^\circ$$ in the xy-plane?

Dhamnekar Winod said:
Hello,What is translation vector t ? and what is x' ?

In the case of z=2*x*y, what is translation vector t and what is x' ? If x' is a new coordinates system, then how is it equal to $B^T(x-t)?$My other queries(if any) will be posted in my next reply.:)

In the case $z=2xy$, we have a translation of $\mathbf t=\mathbf 0$.
For instance $z=2(x-1)(y-2)$ would have an actual translation.

The vector $\mathbf x'$ is your $\mathbf p'=(x', y', z')$.
In this case:
$$\mathbf x'=\begin{pmatrix}x'\\y'\\z'\end{pmatrix}=B^T(\mathbf x-\mathbf t) = \begin{bmatrix}1/\sqrt{2}&1/\sqrt{2}&0\\-1/\sqrt{2}&1/\sqrt{2}&0\\0&0&1\end{bmatrix}\begin{pmatrix}x-0\\y-0\\z-0\end{pmatrix} \implies \begin{cases}x'=\frac x{\sqrt 2}+\frac y{\sqrt 2} \\ y'=-\frac x{\sqrt 2}+\frac y{\sqrt 2} \\ z'=z \end{cases}$$

Re: How $$z=2*x*y$$ is a Hyperbolic Paraboloid in the rotated $$45^\circ$$ in the xy-plane?

Klaas van Aarsen said:
In the case $z=2xy$, we have a translation of $\mathbf t=\mathbf 0$.
For instance $z=2(x-1)(y-2)$ would have an actual translation.

The vector $\mathbf x'$ is your $\mathbf p'=(x', y', z')$.
In this case:
$$\mathbf x'=\begin{pmatrix}x'\\y'\\z'\end{pmatrix}=B^T(\mathbf x-\mathbf t) = \begin{bmatrix}1/\sqrt{2}&1/\sqrt{2}&0\\-1/\sqrt{2}&1/\sqrt{2}&0\\0&0&1\end{bmatrix}\begin{pmatrix}x-0\\y-0\\z-0\end{pmatrix} \implies \begin{cases}x'=\frac x{\sqrt 2}+\frac y{\sqrt 2} \\ y'=-\frac x{\sqrt 2}+\frac y{\sqrt 2} \\ z'=z \end{cases}$$

Hello,

So far we have seen how to compute new coordinates of the same point in $\mathbb R^3$ after rotating the xy-plane by some measures(degree or radians) of angle.

But how to compute new coordinates of the same point in $\mathbb R^3$ if1) We rotate the xy-plane and yz-plane by some measure of angle simultaneously (same radians for both the planes or different radians for each of the two planes).

2)We rotate all the three planes namely xy, yz, xz-planes by some measure of angle simultaneously(same radians for all the three planes or different radians for each of these three planes).

If you or any member of MHB know the correct answer to this question may submit a reply.:)

Re: How $$z=2*x*y$$ is a Hyperbolic Paraboloid in the rotated $$45^\circ$$ in the xy-plane?

Dhamnekar Winod said:
Hello,

So far we have seen how to compute new coordinates of the same point in $\mathbb R^3$ after rotating the xy-plane by some measures(degree or radians) of angle.

But how to compute new coordinates of the same point in $\mathbb R^3$ if1) We rotate the xy-plane and yz-plane by some measure of angle simultaneously (same radians for both the planes or different radians for each of the two planes).

2)We rotate all the three planes namely xy, yz, xz-planes by some measure of angle simultaneously(same radians for all the three planes or different radians for each of these three planes).

If you or any member of MHB know the correct answer to this question may submit a reply.:)

Best is to identify a rotation by its axis and a single angle - the rotation angle around the axis.
We can deduce the axis since it is the vector that $B$ transforms into the same vector.
In the example $z=2xy$ it is a vector in the direction of the positive z-axis.

If you want, we can construct a rotation matrix based on angles with the xy- and yz-planes (1), or the 3 Euler angles (2).
Ultimately we will typically want to reduce them to an identification of the rotation axis and the rotation angle around that axis though.
If you want to know how to do that, I suggest to make that a different question, since that is not really related to the current one.

## 1. What is a Hyperbolic Paraboloid?

A Hyperbolic Paraboloid is a three-dimensional surface that resembles a saddle or a Pringles chip. It is formed by the intersection of a plane with two sets of parallel lines, one set being curved and the other set being straight.

## 2. How is z=2xy a Hyperbolic Paraboloid?

The equation z=2xy represents a Hyperbolic Paraboloid because it satisfies the criteria of having two sets of parallel lines, one set being curved (represented by the xy terms) and the other set being straight (represented by the constant 2).

## 3. What does it mean for the Hyperbolic Paraboloid to be rotated 45° in the xy-plane?

When we say that the Hyperbolic Paraboloid is rotated 45° in the xy-plane, it means that the surface is tilted at a 45° angle with respect to the x and y axes. This rotation can be visualized as the surface being twisted or warped.

## 4. How is z=2xy a Hyperbolic Paraboloid in the rotated 45° in the xy-plane?

The equation z=2xy satisfies the criteria for a Hyperbolic Paraboloid, and when it is rotated 45° in the xy-plane, it forms a twisted or warped surface that resembles a saddle or a Pringles chip.

## 5. What are the applications of Hyperbolic Paraboloids in science?

Hyperbolic Paraboloids have many applications in science, including in architecture, engineering, and physics. They are used in the design of roofs, bridges, and other structures, as well as in mathematical models and simulations in physics and other fields.

Replies
5
Views
2K
Replies
1
Views
2K
Replies
10
Views
3K
Replies
2
Views
2K
Replies
1
Views
5K
Replies
2
Views
3K
Replies
3
Views
1K
Replies
14
Views
4K
Replies
1
Views
1K
Replies
1
Views
3K