Professor Dave here, let’s graph some hyperbolas.
We’ve made it to the last conic section, the hyperbola.
This is the one that extends through the base of one side of the cone all the way through
to the base of the other side of the cone, producing something that at first looks kind
of like a double parabola, with its two separate branches, although the shape is actually quite different.
The definition of a hyperbola is similar to that of an ellipse, where there are two foci,
except that while every point on an ellipse has distances to these foci that give a constant
sum, every point on a hyperbola has distances to the foci that give a constant difference.
The foci sit here, on the outsides of the branches.
Then here we have the two vertices, and between those we can find the center, which sits on
the transverse axis that connects the vertices.
The equation for a hyperbola will be extremely similar to that of an ellipse.
In fact, the only difference is that instead of a plus here, we have a minus, which makes
sense given the difference between distances to the foci being constant instead of the sum.
Everything else will be the same, except that for an ellipse, we always put X squared first
and Y squared second, while the A and B terms could move around.
This time, the A and B terms will stay put, while X and Y can swap spots.
If it’s the X term first, the hyperbola will open up around the X axis.
If the Y comes first, it will open up around the Y axis.
The A, B, and C terms we know will mean pretty much the same things, as well.
A is the distance from the center to a vertex, and C is the distance from the center to a focus.
One difference is that B no longer means much graphically, and we will just use it to solve
for A or C, and this time the equation will be C squared equals A squared plus B squared.
Let’s try an example.
Here’s an equation in standard form, where A squared is sixteen and B squared is nine.
A is the distance to a vertex, so the vertices must be at plus and minus four.
Then, we do C squared equals sixteen plus nine, or twenty-five, so C is five, and the
foci must be at plus and minus five.
And that’s enough information to do a rough sketch of the hyperbola.
If the X and Y terms were reversed, we would have a similar situation, except that now
the transverse axis is vertical, but the distance to the vertices and foci will remain the same.
Just like with ellipses, we can also be given coordinates for vertices and foci and generate
the equation from that.
Vertices on the X axis at plus and minus two and foci at plus and minus five means that
twenty five equals four plus B squared, which makes B squared twenty-one.
So our A squared and B squared terms go here, the X term comes first since it’s horizontal,
and that’s the equation for the hyperbola.
If you’re having trouble drawing the branches of the hyperbola, one thing that can help
is to draw the asymptotes.
These are the lines that the branches approach, getting closer and closer but never touching,
and for a hyperbola that opens around the X axis, these asymptotes are given by the
equations Y equals B over A times X, and Y equals negative B over A times X.
For a hyperbola that opens around the Y axis, the asymptotes are given by the equations
Y equals A over B times X, and Y equals negative A over B times X.
If you draw these asymptotes with dotted lines, they can help you draw the hyperbola a bit
more accurately.
Again, if a hyperbola is not in standard form because this number is not one, we just divide
through by that number to get rid of it.
And finally, hyperbolas can be transformed similarly to ellipses.
If we rewrite as X minus H and Y minus K, the point HK will be the new center of the
hyperbola.
That’s what we need to know about hyperbolas, so let’s check comprehension.