How do we draw the "left and right" contours of a hyperbolic function?

answered 2014-07-09 11:08:52 -0600

Anna
101 ●3 ●5

updated 2014-07-09 11:16:37 -0600

Mark
166 ●1 ●5 ●12 https://sites.google.com/...

Hey so an easy way to show the graphs is by first putting the equation into wolfram and then copying the link and pasting it into the "image" thing above. Below is a graph of $ 4x^2 -9y^2=1 $

And I'm pretty sure you can get the other contours by setting z as a negative number. Below is a graph of $ 4x^2 -9y^2=-1 $

I'm not sure about any of this though so feedback and/or other ideas would be very much appreciated!

Comment: I love WolframAlpha! The key issue by hand, though, is finding the intercepts.

answered 2014-07-09 15:31:51 -0600

Christina
807 ●18 ●35 https://sites.google.com/...

This may be completely wrong because this is difficult for me to do as well, but I think what we need to do here is to set the constant (which is the $f(x,y)$) equal to several different values in order to sketch the contours. As you stated above, setting $f(x,y) = 1$ will only give you one hyperbola to draw. But then when we set the constant to other values, we get more and more hyperbolas to sketch.

So what happens when we set the constant to a negative value? This would change the above equation to $9y^2 - 4x^2$, which would then give us the other hyperbolas to draw.

I am a bit rusty, but to graph a hyperbola it helps to get the right form. So to get rid of the coefficients, we could set our equation equal to 36 and -36. That would give us the following:$$\frac{x^2}{9} - \frac{y^2}{4} = 1$$ and $$\frac{y^2}{4} - \frac{x^2}{9} = 1$$ If I am correct in my thinking, that would give us our first vertices (intercepts) of (0,3), (0,-3), (2,0), and (-2,0)....from there we could set our equation equal to other values to get more contours.

Does this sound right? It has been a while!