Ellipses - At A Glance

Find the center, vertices, and foci of .

The equation of an ellipse is .

The center is at (h, k), which is (-3, 3) in this case. It's easy to see that a = 10, but where did b go? It's still there, underneath y, we just can't see it because it is 1. Very sneaky, b.

We can already find all four vertices for the ellipse. a is the horizontal distance from the center to two of the vertices, and b is the vertical distance for the other two. x and a go together, and so do y and b. This gives us:

(-3 ± 10, 3) and (-3, 3 ± 1)

(-13, 3), (7, 3), (-3, 2), and (-3, 4)

The ellipse axis parallel with the x-axis is definitely bigger than the one parallel to the y-axis. That makes a = 10 the length of the semi-major axis, and the ellipse as a whole is horizontal.

That's all we need to know to find f, the distance to the foci. We're on to you, f, and your accomplice, f². To find him, we take the square of the semi-major axis and subtract away the square of the semi-minor axis.

f² = a² – b²

= 100 – 1 = 99

The foci fall on the major axis, so we'll add and subtract f from the x-coordinate of the center.

(-3 ± 9.95, 3)

(-12.95, 3) and (6.95, 3)

Oh boy, now we've done it. Actually, we've solved the problem. That's the thing we did. Nothing wrong with that.

Find the center, vertices, and foci of  .

Let's get the easy stuff out of the way first. The center is (0, 0), which really fits the bill as "easy to work with," and a = 9 while b = 11.

The larger semi-axis is b = 11, so we have a vertical ellipse. Our vertices are at (-9, 0), (9, 0), (0, -11), and (0, 11). All we need now are the foci.

f² = b² – a²

The larger semi-axis goes first.

f² = 121 – 81

= 40

The foci fall on the major axis, so they'll boogie up and down from the center. They are at (0, -6.32) and (0, 6.32).

Find the equation of the ellipse with vertices at (-3, 4) and (5, 4), and foci at (-1, 4) and (3, 4).

We seem to be missing a center here. That's kind of important. We have two vertices and both foci, though, and that will be enough to find the center. The foci in particular are helpful. Both of them have a 4 in the y-coordinate. Our center will need a 4 there too, to line up with them.

If y = 4 is fixed, that creates a horizontal line. This is the major axis, then, and so we have a horizontal ellipse. We should find a > b once we get that far.

Back to the center. What's the x-coordinate? It has to be in between the two foci, or the two vertices. Sticking with the foci, halfway between -1 and 3 is +1. Therefore, the center is (1, 4). We can also see that f must be 2.

Now we need a and b. The vertices we know only vary in the x-coordinate, so they must be associated with x. We'll use them to find a. Starting from x = 1, from the center, each vertex is 4 away. That's our a.

Our semi-major axis is a, so the formula to get to b is:

f² = a² – b²

4 = 16 – b²

Don't forget to use the squares here. You don't want to upset Pythagoras's ghost, now would you?

b² = 12

We're sticking everything in the equation for an ellipse, and then we are out of here.