How to sketch $-3x^2 - 8xy + 3y^2 = 1$

Question

The equation is as follows:

$$-3x^2 - 8xy + 3y^2 = 1$$

How to specify the axis of the given curve? How to as accurately as possible draw a curve defined by this equation?

Answer 1

First curve tracing. The invariant $I_2 = 8^2- ( -9) >0 $, it is a hyperbola. Expect two blunt nose shaped curves. Find values of x for $y=0$ and y for $ x= 0.$ to trace it approximately.

Put $0$ in place of $1$ and factorize, getting a quadratic in $ y/x$, yielding two straight line asymptotes (equations as a product of two asymptotic straight lines like $(y-m_1 x)(y-m_2 x)=0. $

Answer 2

Part of it is recognition. You can recognize that all the terms in this polynomial equation are quadratic (or lower degree), so that automatically means you have a conic section.

Generally, there may be linear terms, but a substitution $x\to u+a, y\to v+b$ can be used to eliminate linear terms, and work in a shifted $uv$-axis system. Here, we get to skip that step.

The next step is to write the equation using its bilinear form matrix: $$\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}-3&-4\\-4&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=1$$

This matrix's eigenvalues and eigenvectors tell you the behavior of the conic section. Here, the eigenvectors are $v_{-5}=\begin{bmatrix}2\\1\end{bmatrix}$ and $v_{5}=\begin{bmatrix}-1\\2\end{bmatrix}$. This tells us that the conic sections primary axes are these two orthogonal directions. And in a rotated axis system using $s$ as the variable along $\begin{bmatrix}2\\1\end{bmatrix}$, and $t$ as the variable along $\begin{bmatrix}-1\\2\end{bmatrix}$, that the equation is $$-5s^2+5t^2=1$$ which, if you are familiar with very basic hyperbolas, tells you that you have a hyperbola. Along the $s$-axis, there is no intersection with the curve. Along the $t$-axis, there are vertices at $\pm\sqrt{1/5}$. And the ratio of $-5/5$ tells you the asymptotes are orthogonal. That means they must have angles from the origin halfway between the angles that make the axes.

So here is what that all tells us:

Answer 3

Take the derivatives with respect to $x$ and $y$ and subtract them

$$ f(x,y) = 3 x^2-8 x y+3 y^2-1 = 0$$

The solution: $$ \frac{\partial f(x,y)}{\partial x} - \frac{\partial f(x,y)}{\partial y} = 0 $$ $$ (6 x-8 y)-(6 y- 8 x) =0 $$ $$ 14 (x-y) =0 $$ $$ y=x $$

Also where $\frac{\partial f(x,y)}{\partial x}$ and $\frac{\partial f(x,y)}{\partial y}$ meet is the center.

Also note that this works with other hyperbolas, but does not work with ellipses. You need to add the partials to get the axis of an ellipse. I don't know why.