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.
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.