Let's dive right in. Here's the wonder equation that drives an ellipse:
We have the familiar variables h and k returning, and this time they point right at the center of the ellipse. Both x and y are squared in this equation, and the whole thing sums up to 1.
What good is half an axis? It's all good. That's because a and b give us the lengths of the semi-major and semi-minor axes. Here "semi" is used in the sense of half, not a big rig. A semi-axis goes from the center to one of the vertices.
a is under x, and so it is the horizontal distance from the center, while b, underneath y, gives the length of the vertical axis. In a question that every inequality sign recognizes, we have to ask, "Which one is bigger?" The larger denominator goes with the major axis, and the smaller one goes with the minor axis.
At this point, we can draw a pretty wicked ellipse. We can even draw a wicked face on it, if we want, and it will be pretty accurate. So we guess we're done, right? Sorry, but no dice. We still need to track down the two foci. They're located on the major axis, but where?
Will having the foci help us with sketching the graph? No, although if we wanted to be super accurate we would need them. Mathematicians, though, love knowing the foci. It's like catnip for them. We could watch them play with the foci all day.
We need to use an equation to find the distance from the center to one of the foci. It's a little specific, though, so bear with us. Unless you're actually a bear, in which case please don't eat us.
A life without f is a life without foci. We want f, if only to keep the mathematicians happy. The secret to finding it is knowing that the 
is equal to 
. The diagonal is the same as the semi-major axis. We don't know why, but we suspect those mathematicians had something to do with it.
We have a right triangle sitting in the middle of our ellipse, so we can use the Pythagorean Theorem to find f2. For only working on right triangles, Pythagoras sure has a way of getting around. Where will he show up next?
Anyway, in a horizontal ellipse, we can see that f2 = a2 – b2. For a vertical ellipse, it is f2 = b2 – a2. For any ellipse ever, it is f2 = (square of the semi-major axis) – (square of the semi-minor axis).
With the focus distance under our belt, we now have all the pieces needed to work with ellipses—the center, major and minor axes, vertices, and foci. If you don't have a belt, guess you can't solve any of the problems in this section. Oh fine, you can borrow one of ours.
Sample Problem
Graph the ellipse 
, including its foci.
Our checklist for graphing an ellipse includes: the center, the lengths of the semi-major and semi-minor axes, 2 eggs, and 2 foci. The only things that look hard to find in this equation are the eggs, so we'll make a separate trip to the store for those.
Checking h and k, next to x and y, suggests that the center is located at (1, 0). Well, it does more than suggests it; it demands it. It better be the center, or h and k are going to have words with us. Nobody wants that.
The denominators beneath x and y are a2 and b2. Taking the square roots of them gives us a = 5 and b = 3, and these are the lengths of the semi-axes. The semi-axis under x is larger, so we have a horizontal ellipse—think of it like a sub sandwich, instead of burrito-shaped.
Now to hunt down the foci. Maybe they're between the tomato and lettuce? Wait, the ellipse is only like a sub; it isn't actually one. Guess we better use the formula for f2 instead:
f2 = (semi-major axis)2 – (sem-minor axis)2
= 25 – 9 = 16
f = 4
f is a distance, so it has to be a positive real number. If we had tried subtracting 25 from 9, instead of doing it the right way, getting a negative number would have been a hard reality check.
Anyway, now we can start graphing. Starting at the (1, 0), we go left and right 5 for the horizontal vertices, and up and down 3 for the vertical ones. The foci are tag-alongs that go along with the major axis, so they'll go horizontal too, by f = 4. Putting it all together we get:
That doesn't look much like a sub sandwich at all. Good thing we didn't try to take a bite of it when we had the chance.
Sample Problem
Graph the ellipse 
, including its foci.
Whoa there, partner. Before we go rushing off to graph this puppy, take another look at it. This isn't a proper ellipse until it equals 1. Just like it isn't a proper party until Shmoop arrives. Let's divide everything by 100:
Now we're in business. And that business is the problem solving business. The center is sitting pretty at (2, -2), while a is 2 and b is 7. This time, b > a, so semi-major axis is with y. That's a vertically oriented ellipse. We'd get excited about its tall, burrito shape, but we already know we won't get to eat it.
Let's get f now, so we can get to graphing. The larger axis is b, so we subtract a2 from b2.
f2 = 49 – 4 = 45
Keeping the radical while we try to graph sounds like an exercise in frustration and sadness. We're just going to round that out to 6.7. But at least we have all of the pieces in place to start graphing.
There we go. Graphing a vertical ellipse isn't very different from a horizontal one. The only big difference is what kind of sauce we put on it. Actually, that's burritos and sandwiches again.
Sample Problem
Find the equation of the ellipse with one focus at (1, 2), one vertex at (1, 3), and a center of (1, -1).
Sometimes, we won't start with an equation, but with some of the parts of an ellipse. We have to work backwards without bumping into anything.
To write out the equation of an ellipse, we need h, k, a, and b. Other letters of the alphabet can lend a hand, but no one else is actually in the band. We're given the center, so we already have h and k.
Comparing the vertex to the center will give us a or b. We see that there's a difference of 4 in the y value, so b = 4. So far, so good. We can find f in the same way from the focus; it has a y difference of 3. Together, they can find a.
Here's where we hit a snag. Is b the semi-major or semi-minor axis? We don't know if we should use f2 = b2 – a2, or f2 = a2 – b2. Or maybe we should call the cops and have them bring out the tracking dogs to find a?
Put down the phone, that wasn't a real choice. Instead, look at the three points we were given at the start of the problem. (1, 2), (1, 3), and (1, -1)—all three of them are lined up on the same x-coordinate. We don't move left or right to move between the focus, the center, and the given vertex; we move up and down. The major axis must be vertical, so b must be the semi-major axis.
That means 9 = 42 – a2. A little rearranging and solving gets us 
. That's it, we're done. We just need to put all of our information into the formula for an ellipse.
Ellipse found. Now we can call off the search teams and stop putting its description on milk cartons.