orthogonal diagonalization to sketch equation: $5x^2-24xy-5x=13$

Question

to sketch this i wrote the equation down in the form:
$X^TAX=13$ where $X^T=[x\:\:y]$ and $A=\begin{bmatrix} 5&-12\\ -12&-5 \end{bmatrix} $.
Then, $A$ is orthogonally diagonalizable so there exist a matrix $D$ and an orthogonal matrix $Q$ such that $Q^TAQ=D$. orthogonally diagonalizing the matrix gives me : $ D= \begin{bmatrix} -13&0\\ 0&13\\ \end{bmatrix}$ and $ Q=\begin{bmatrix} \frac{2}{\sqrt{13}}&-\frac{3}{\sqrt{13}}\\ \frac{3}{\sqrt{13}}&\frac{2}{\sqrt{13}} \end{bmatrix} $ .
Then i can make new coordinates such that $X=Q\:X'$, where $(X')^T=[x'\:\:y']$.
$(X')^TDX'=13$, so multiplying out these matrices gives the equation in terms of the new coordinates: $(y')^2-(x')^2=1$ which i think is an hyperbola that cuts $y'$ at 1 and -1, so what i thought now is that in terms of the original $x$ and $y$ it should look like an hyperbola rotated clockwise by roughly 56 degrees, since $X$ is just $X'$ rotated anti-clockwise by $Q$. But when i actually sketch the curve with a computer i get the hyperbola rotated anti-clockwise. why is that? Can someone point out where the mistakes are please?

Answer 1

First, your calculations are correct. I think your difficulty is that there are two ways of looking at these problems and you are mixing them up.

One possibility is to say that we are looking at a curve and identifying it by changing the coordinates in order to eliminate the $xy$ term. So it is the same curve in the same location in the plane, but we are describing it in terms of new variables $(x',y')$, giving the equation $(y')^2-(x')^2=1$. In this interpretation the curve has not rotated at all, either clockwise or anti, it's the axes that have rotated. A unit vector in the $x'$ direction is $$X'=\pmatrix{1\cr0\cr}$$ and in original coordinates this is $$X=Q\pmatrix{1\cr0\cr}=\frac1{\sqrt{13}}\pmatrix{2\cr3\cr}\ ,$$ which is about $56^\circ$ anticlockwise from the $x$ axis. I think this is where you have misinterpreted the computer output - you think the hyperbola has rotated anticlockwise, but actually it hasn't moved at all, it's the axes which have rotated. If you look at a sketch of the curve with both $(x,y)$ and $(x',y')$ axes on the same diagram, it may help you to understand what going on.

The second possibility is to imagine that the curve does in fact rotate. In this case you obtain a geometrically congruent curve (but not the same curve, i.e., not the same points) located differently in the plane. Its equation will be $y^2-x^2=1$ (note: not $(y')^2-(x')^2=1$), and if you think about it a bit this means that the curve will rotate in the opposite direction from what the axes did in the previous interpretation, so, $56^\circ$ clockwise.

It's a bit tricky to sort out what is happening with the two ways of looking at essentially the same problem, but I hope this helps.