Various ways to intuitively draw the graph of $x^2 -y^2 = $constant?

Question

How can I intuitively draw the graph of $x^2 - y^2 = $ constant ? like Intuitively I can draw the graph of $xy =$ constant, like if $x$ increases then $y$ has to decrease so as to be constant and we get a rectangular hyperbola, but how do i think in this case of $x^2 - y^2 = $ constant ?

Answer 1

Questions of "intuition" can be fairly subjective, so here is an example of my thought process when I try to visualize $x^2-y^2=\text{constant}$.

First, rewrite it as $y^2=x^2-\text{constant}$. Notice that for any pair of solutions $(x_0, y_0)$, that $(-x_0, y_0)$, $(x_0, -y_0)$, and $(-x_0, -y_0)$ are also solutions. So the graph must be symmetric about the x and y axes.

Also notice that for very large x, we have $y/x \approx \pm 1$. So the further from the origin you get, the more it looks like the lines $y=\pm x$. This can be more formalized with limits if you don't like my hand-wavey explanation.

Finally, look at the solutions when $y=0$ and when $x=0$. If the constant is negative, we have $y=\pm \sqrt{\text{-constant}}$ when $x=0$. Or if the constant is positive, then we have $x=\pm \sqrt{\text{constant}}$ when $y=0$. Assuming the first case without loss of generality, we can see that the graph will never cross one of the axes, because this point corresponds to the minimum/maximum above/below this axis.

Put all of this information together, and you have what amounts to a hyperbola.

Answer 2

you have $$(x+y)(x-y)=c$$

So when $(x+y)$ is very large, $(x-y)$ is very small. Therefore $x\simeq y$ for large $x,y$

Therefore the line $y=x$ is an oblique asymptote.

Conversely so is $y=-x$

You also see what happens when either $x$ or $y$ is zero...

Answer 3

$x^2-y^2$ can be factored into $(x-y)(x+y).$ As $y$ goes up in one factor, it will go down in the other factor. As $x$ goes up in one factor, it will also go up in the other factor.

If you want to try thinking of it a different way, try to picture the surface $x^2-y^2=z$. A plane at height $z$ cuts through this, and that's what you are seeing when the constant is $z$. You know what $x^2=z$ looks like, so you can plot this over and over again, adding different values of $-y^2,$ to build up the 3D function.

Here is a picture from Wolfram Alpha: plot of x^2-y^2=z

Answer 4

This may not be what you want but I'll add a slightly different perspective. Please ignore if not relevant.

Consider that a circle centre $(0,0) $ with radius $r$ has equation $x^2+y^2=r^2$ in the cartesian coordinate system. Now imagine the mapping $y \mapsto iy$ the equation of a circle becomes $x^2+(iy)^2=r^2\implies x^2-y^2=r^2$

I like to think of this as a reason why trig and hyperbolic functions are so intertwined.

To answer the question it is a circle under the mapping.

Answer 5

Make the substitutions $u=x+y,$ $v=x-y. $ This change of variables has the effect of rotating the coordinate axes by $45^{\circ}$ about the origin, and turns the equation into $uv=\text{const.},$ which you already know how to draw.