Hyperbola given by the midpoint of two LI vectors

Question

Here is the problem:

Let u and v be two LI vectors in $\mathbb{R}^3$.

a) Prove that the area of the triangle with sides $\lambda \overrightarrow{u}$ and $\frac{1}{\lambda}\overrightarrow{v}$ doesn't depend on $\lambda$.

b) Prove that the midpoints $M_{\lambda}$ of $\lambda u$ and $\frac{1}{\lambda}v$ lay on a Hyperbola. Find its vertices and focus, in terms of $u$ and $v$.

Item a) it's easy. If we use the trigonometric formula for the area of triangle, we can conclude that the area is given by $\frac{||u||\cdot ||v||\sin\theta}{2}$.

But I have no idea of how to start item b). I already used a software to see that this midpoint really lays on a hyperbola. I tried to look just inside the plane formed by this two vectors, but I can't figure out how to get the standart form ($\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$) of the hyperbola. I thought of using the definition of a hyperbola, the one with the geometric locus: $|d(P,F_1)-d(P,F_2)|=2a$, but we would need to find the focuses first.

Another approach that a could think of was to look this problem in 2 dimension and to stablish a coordinate system so that the x-axis would be the bisector of $\lambda \overrightarrow{u}$ and $\frac{1}{\lambda}\overrightarrow{v}$, but I couldn't translate that into algebra.

Here's some pics I used to verify if this problem does make sense:

Them I run some values for $\lambda$ and traced the midpoint:

Software: Geogebra

In this particular case I took $u=(2,2)$ and $v=(1,0)$

Any hint is enough.

thanks in advance.

Can you please edit this question to include its source? (Textbook/PDF/exam question etc.). Also include your background and how much you know about this subject in particular (e.g. about formulas for a hyperbola and so on). — Sarvesh Ravichandran Iyer, Nov 08 '23 at 12:33
If you can attach photos of the hyperbola or describe the software you used/ the simulation you got/ the code you wrote, that would also be nice. — Sarvesh Ravichandran Iyer, Nov 08 '23 at 12:40
@SarveshRavichandranIyer, I'm a math teacher who is almost getting a Master Degree in Mathematics. — Igor Souza, Nov 08 '23 at 16:39
Thank you so much Igor. I am confident that you can understand whatever I'm about to write as an answer then.(It will take some time) — Sarvesh Ravichandran Iyer, Nov 08 '23 at 17:19
$(x,y)=(\lambda\frac{x_1}2+\frac1{\lambda}\frac{x_2}2,\lambda\frac{y_1}2+\frac1{\lambda}\frac{y_2}2)$ and eliminating $\lambda$ we get the equation $$4y_1y_2x^2+(-4y_1x_2-4x_1y_2)xy+4x_1x_2y^2+y_1^2x_2^2-2x_1y_1x_2y_2+x_1^2y_2^2=0$$ or $$4 (x_1 y - x y_1) (x_2 y - x y_2) + (x_2 y_1 - x_1 y_2)^2=0$$ which is of the form $\text{asymptote}_1\text{asymptote}_2=\text{constant}$ as are all hyperbolas. — Jan-Magnus Økland, Nov 08 '23 at 18:05
@Jan-MagnusØkland Very interesting. Do you know any source for me to study about this last form of hyperbolas? Once proved it is in fact a hyperbola, I want to find some aspects of that hyperbola, like vertices and focuses. — Igor Souza, Nov 08 '23 at 18:34
@IgorSouza you can prove it by recognizing that the standard form is $(\frac{x}{a}-\frac{y}{b})(\frac{x}{a}+\frac{y}{b})=1.$ — Jan-Magnus Økland, Nov 08 '23 at 22:21
I have been unable to solve this question using the idea I had in mind. I apologize for the same. — Sarvesh Ravichandran Iyer, Nov 09 '23 at 12:18

Hosam Hajeer · Answer 1 · 2023-11-08T22:34:46.853

Let the midpoint be

$ r = [x, y]^T = \dfrac{1}{2} ( \lambda u + \dfrac{1}{\lambda} v ) $

Then,

$ r = \dfrac{1}{2} [ u, v] w \hspace{20pt}(1) $

where $ w= [ \lambda , \dfrac{1}{\lambda} ]^T $

Note that $ w^T Q_0 w = 1 \hspace{20pt}(2)$

where

$ Q_0 = \begin{bmatrix} 0 && \dfrac{1}{2} \\ \dfrac{1}{2} && 0 \end{bmatrix} $

From $(1)$ , we have

$w = [ u, v]^{-1} (2 r ) $

Substitute this into $(2)$, you get,

$ 4 r^T [ u, v]^{-T} Q_0 [u, v]^{-1} r = 1 $

Since $Q_0$ has one positive eigenvalue and one negative eigenvalue then so does the matrix $Q = 4 [u, v]^{-T} Q_0 [ u, v ]^{-1} $, and since it is a symmetric matrix it can be diagonalized as follows

$ Q = 4 [u, v]^{-T} Q_0 [u, v]^{-1} = R \ D \ R^T $

where $R$ is a rotation matrix, and $D$ is a diagonal matrix. Hence, the equation now becomes

$r^T R D R^T r = 1 $

Take $D = \text{diag}\{ D_{11} , D_{22} \}$ with $D_{11} \gt 0 $ and $D_{22} \lt 0 $. Also take $r' = [x',y']^T = R^T r $ (i.e. $r = R r'$), then

$ {r' \hspace{2pt}}^T D \ r' = 1 $

which reads,

$ D_{11} (x')^2 + D_{22} (y')^2 = 1 $

Considering the signs of $D_{11}$ and $D_{22}$ , then this is a hyperbola in standard position/orientation with semi-major axis equal to $\dfrac{1}{\sqrt{D_{11}}} $ and semi-minor axis equal to $ \dfrac{1}{\sqrt{- D_{22}}} $. It follows that $r$ is the same hyperbola as $r'$ but rotated by an angle $\theta$ which is obtained from $R$, because $R$ is equal to

$ R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $

Taking your example, you have $u = [2, 2]^T $, $ v = [1, 0]^T $

Then

$ [u, v]^{-1} = \begin{bmatrix} 2 && 1 \\ 2 && 0 \end{bmatrix} = - \dfrac{1}{2} \begin{bmatrix} 0 && -1 \\ -2 && 2 \end{bmatrix} $

Then

$Q = 4 [u, v]^{-T} Q_0 [u, v]^{-1} = \begin{bmatrix} 0 && -2 \\ -1 && 2 \end{bmatrix} \begin{bmatrix} 0 && \dfrac{1}{2} \\ \dfrac{1}{2} && 0 \end{bmatrix} \begin{bmatrix} 0 && -1 \\ -2 && 2 \end{bmatrix} = \begin{bmatrix} 0 && 1 \\1 && -2 \end{bmatrix}$

Diagonalzing $Q$, we get

$ Q = R D R^T $

with

$ D = \begin{bmatrix} \sqrt{2} - 1 && 0 \\ 0 && - 1 - \sqrt{2} \end{bmatrix} $

$R = \begin{bmatrix} 0.923879533 && - 0.382683432 \\ 0.382683432 && 0.923879533 \end{bmatrix} $

This $R$ is a rotation matrix by an angle $\theta = \text{ATAN2}( 0.923879533 , 0.382683432 ) = \dfrac{\pi}{8} = 22.5^\circ$

The semi-major axis length is

$ a = \dfrac{1}{\sqrt{\sqrt{2}-1}} =\sqrt{\sqrt{2}+1} = 1.55377397403 $

and the semi-minor axis length is

$ b = \dfrac{1}{\sqrt{ 1 + \sqrt{2}} } = \sqrt{\sqrt{2} - 1} = 0.64359425291 $

The distance of each of the foci from the center which is at the origin, is

$ c = \sqrt{a^2 + b^2} = \sqrt{ 2 \sqrt{2} } =1.68179283051 $

The vertices of this hyperbola are at

$ \pm a (\cos \theta, \sin \theta) = \pm (1.4355, 0.5946) $

This results compare very well with the graph you've presented in your question.

Hyperbola given by the midpoint of two LI vectors

1 Answers1