The Ellipse

### The ellipse

An ellipse is one of the so-called conic sections, figures that result from slicing a right-regular cone in one of three ways. The figure below shows how that works for an ellipse. There is only one way to form a closed figure by slicing a cone, and that's to do it without intersecting the base. If that cut is made parallel to the base we end up with a circle, which is just a special case of the ellipse. Any other such cut is an ellipse.

Ellipses are very important in many areas of science and technology. They arise quite naturally in many areas. One is planetary orbits. The orbit of any planet is an ellipse with the body being orbited located at one of the two foci of the elliptical path. All celestial bodies – planets, stars, comets, asteroids, &c. – orbit in elliptical paths, though one body can affect the orbit of another by perturbing it slightly.

### Anatomy of an ellipse: terminology

1 / 10 Conic section
2 / 10 An ellipse
3 / 10 Vertices
4 / 10 Major axis & symmetry
5 / 10 Minor axis & symmetry
6 / 10 The foci
7 / 10 Major axis
8 / 10 Minor axis
9 / 10 Dimensions
10 / 10 Constant distances

Click/tap on the blue arrows in the figure to learn about the anatomy and terminology of the ellipse. You'll have to be familiar with all of the terms and measurements in order to understand what's to follow

#### Focus / foci

These are points along the major axis of an ellipse that determine how elongated or eccentric it is. They are analogous to the center of a circle, and in fact when the foci (plural of focus, pronounced fo'·sy) of an ellipse are at the same point, the ellipse is a circle. These are labeled f1 and f2. The distance between a focus and the center of the ellipse is labeled c.

#### Major & minor axes

The major axis is the longest line of mirror symmetry that can be drawn through an ellipse. It contains both foci. The minor axis is the shortest axis of mirror symmetry.

#### Dimensions of an ellipse

a = one half of the length of the major axis (sometimes called the semi major axis)

b = one half of the length of the minor axis (sometimes called the semi minor axis).

c = the distance between the center of an ellipse and each focus. The distance between the foci is 2c.

d1, d2 are the distances from each focus, f1 and f2, to any point on the ellipse.

#### d1 + d2 = 2a,

always, for any ellipse.

### Drawing an ellipse

Play this clip to show one way that an ellipse can be drawn.

Imagine that we place two tacks (the foci) in a piece of paper. Now tie a loop of string and loop it around the two tacks. Make a triangle of the loop by stretching it out with a pencil, the pencil being the third vertex. Now move the pencil with the string taut and trace out the resulting figure, just like in the animation.

Now the loop of string has a constant perimeter, and the distance between the foci is constant, therefore the sum of the lengths of the strings from each focus to the pencil are the same – the definition of an ellipse.

Notice also that when the distance between the two foci is zero, the loop of string becomes a line, the radius of a circle.

When the distance between the foci of an ellipse is reduced to zero (2c = 0), the ellipse is a circle. A circle is a special case of the ellipse.

### Geometry of an ellipse

One way to approach the geometry of the ellipse is to start with a circle. At any point on a circle centered at the origin, the coordinates (x, y) satisfy the equation

$$x^2 + y^2 = r^2$$

where r is the radius of the circle. Here's a drawing of that. Two points are highlighted on the circle and the relationship between their x- and y-coordinates to the radius is in the right triangles and the Pythagorean theorem. Now an ellipse can be considered to be a circle with two separate characteristic radii, a major one of length a and a minor one of length b. We just have to figure out the analogous relationship between the "radius" at any point on the ellipse and the three dimensions that characterize it, a, b and c. Here's that diagram again: We need to find a formula for the location of any point (x, y) on the ellipse, given that we know that the sum d1 + d2 is constant.

We begin by trying to find the lengths of d1 and d2 by drawing two right triangles like in this graph. On the right the length of the lower leg is c - x, and on the left it's c + x. So using the Pythagorean theorem, we find the lengths of d1 and d2:

\begin{align} d_1 &= \sqrt{(x + c)^2 + y^2} \\[5pt] d_2 &= \sqrt{(x - c)^2 + y^2} \end{align}

Now take a look at the figure below, in which the point P has been moved to a location along the major axis of the ellipse (the overlapping lines d1 and d2 have been moved apart a little so you can see both of them). If a is the dimension shown (the semi-major axis length), then the length of d1 in this position is a - c, where c is the distance from a focus to the center. Likewise, the length of d2 is a + c. Adding those, we get

\begin{align} a + c + (a - c) &= 2a, \; \text{ so} \\[5pt] d_1 + d_2 &= 2a \end{align} The rest of the derivation (below) involves a fair bit of algebra consisting of simple steps, but there's a lot of it. It's here if you're curious about it.

Before we get to that we need to do just one more thing. Given the result that d1 + d2 = 2a, if we draw d1 and d2 to a point P on the minor axis, the resulting equilateral triangle gives us the Pythagorean relationship

$$a^2 = b^2 + c^2$$ In what follows, we'll need that relationship, rearranged to

$$b^2 = c^2 - a^2.$$

#### The derivation

We begin here:

$$d_1 + d_2 = 2a$$

The distance formula, or equivalently an analysis of the right triangles in the figure above, gives us the lengths of d1 and d2:

\begin{align} d_1 &= \sqrt{(x + c)^2 + y^2} \\[5pt] d_2 &= \sqrt{(x - c)^2 + y^2} \end{align}

Adding those expressions and setting them equal to 2a gives

$$\sqrt{(x - c)^2} + \sqrt{(x + c)^2 + y^2} = 2a$$

Now let's rearrange that equation slightly (moving the second radical to the right side by subtraction) to give:

$$\sqrt{(x - c)^2} = 2a - \sqrt{(x + c)^2 + y^2}$$

Now we square both sides to begin getting rid of the square roots. I say begin because the square on the right will result in a cross- or mixed term containing a root, but we'll get to that later.

$$\left[\sqrt{(x - c)^2}\right] = \left[ 2a - \sqrt{(x + c)^2 + y^2}\right]$$

The square on the left is obvious, and we expand (FOIL) the one on the right to give:

$$(x - c)2 + y^2 = 4a^2 - 4a \sqrt{(x + c)^2 + y^2} + (x + c)^2 + y^2$$

Expanding the squared binomials (x - c)2 and (x + c)2, and canceling terms on both sides of the equal sign like this:

$$\require{cancel} \cancel{x^2} - 2cx + \cancel{c^2} + \cancel{y^2} = 4a^2 - 4a \sqrt{(x + c)^2 + y^2} + \cancel{x^2} + 2cx + \cancel{c^2} + \cancel{y^2}$$

gives us a simpler expression. We'll rearrange it just a bit, keeping the root on the left and the other terms on the right, and simplifying a bit further by dividing through by 4:

$$\cancel{4} a \sqrt{(x + c)^2 + y^2} = \cancel{4} a^2 + \cancel{4} cx$$

The result is

$$a \sqrt{(x _ c)^2 + y^2} = a^2 + cx$$

Now we can again square both sides to finally get rid of the last of the roots. That happens because we're no longer squaring a binomial:

$$\left[ a \sqrt{(x + c)^2 + y^2} \right]^2 = \left[ a^2 + cx \right]^2$$

The result is

$$a^2 [(x _ c)^2 + y^2] = a^4 + 2a^2 cx + c^2x^2$$

We now expand the squared binomial (x + c)2 on the left:

$$a^2 [x^2 + 2cx + c^2 + y^2] = a^4 + 2a^2cx + c^2x^2$$

and multiply through by $a^2,$ canceling terms that again appear on both sides of the equal sign:

$$a^2 x^2 + \cancel{2 a^2 cx} + a^2 c^2 + a^2 y^2 = a^4 + \cancel{2a^2 cx} + c^2 x^2$$

We're getting closer. If we take the result,

$$a^2x^2 + a^2c^2 + a^2y^2 = a^4 + c^2x^2$$

and rearrange it slightly by moving c2x2 to the left and a2c2 to the right, both by subtraction, we get:

$$a^2x^2 + a^2y^2 - c^2x^2 = a^4 + a^2c^2$$

we can then factor an x2 from the left and an a2 on the right to find a binomial in common to both sides:

$$x^2(a^2 - c^2) + a^2y^2 = a^2(a^2 - c^2)$$

We can take that expression and divide both sides by a2(a2 - c2), canceling as we can. That's going to give us 1 on the right, just what we're looking for. It looks like this:

$$\frac{x^2 \cancel{(a^2 - c^2)}}{a^2 \cancel{(a^2 - c^2)}} + \frac{\cancel{a^2} y^2}{\cancel{a^2}(a^2 - c^2)} = \frac{\cancel{a^2(a^2 - c^2)}}{\cancel{a^2(a^2 - c^2)}}$$

The result is getting closer to what we're looking for:

$$\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$$

Now if we remember that b2 = a2 - c2, we can make the direct substitution to get our well-known formula for an ellipse centered at the origin with major axis 2a and minor axis 2b:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

#### The Ellipse

The equation of an ellipse centered at (0, 0) with major axis a and minor axis b (a > b) is

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

If we add translation to a new center located at (h, k), the equation is:

$$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$$

The locations of the foci are (-c, 0) and (c, 0) if the ellipse is longer in the x direction, and (0, -c) & (0, c) if it's elongated in the y-direction, $c^2 = a^2 - b^2.$

#### Exploring the a & b parameters

You can explore how the a and b parameters affect the shape of an ellipse centered at the origin by using this widget.

Move the sliders to change a and b; recall that the larger of the two is always called a, so depending on the shape of the ellipse (elongated left-to-right or up-to-down), one slider corresponds to a, the other to b.

At the crossing point, where a = b, the ellipse becomes a circle, which is really just a special case of the ellipse. Remember, more things are the same in math and science than are different!

Below we'll see how we can apply a rotational transformation to the ellipse to tilt its axes of symmetry with respect to the graph axes.

### Example 1

Sketch the graph of the ellipse described by $\frac{x^2}{2} + y^2 = 3$

Solution: The equation looks roughly like that of an ellipse, but our goal will be to convert it into the correct form so that we can just read off the geometric dimensions of our ellipse and sketch it.

$$\frac{x^2}{2} + y^2 = 3$$

To start, we'll divide by 3 on both sides to get a 1 on the right: Now the root of the larger dimension goes with x and we'll call it a, where 2a is the length of the major axis. The root of the smaller dimension is b, twice the length of the minor axis.

For an ellipse, we showed above that

\begin{align} c^2 &= a^2 - b^2 \\ &= 6 - 3 = 3 \end{align}

So c, the distance between the center of the ellipse and each focus is:

$$c = \sqrt{3} \approx 1.7$$

This ellipse is centered a the origin, so now we have all of the information we need to sketch it.

Begin by drawing a box of dimensions 2a × 2b centered at the origin. The vertices of the ellipse will be tangent to that box. Then sketch in the foci along the major axis, about 1.7 units from the origin. Finally, sketch in the outline of the ellipse. This "box method" of drawing an ellipse from its formula is pretty straightforward. Below we'll do an example of an ellipse elongated along the y-axis and of one displaced from the origin.

### Example 2

Sketch the graph of the ellipse described by $\frac{x^2}{4} + \frac{y^2}{9} = 1$

Solution: This figure is already in in the standard form for an ellipse centered at the origin, so we can just read off what a and b are: a = 2 and b = 3. Notice that a in this case is tied to the y-dimension. That's our convention: a is always the longest axis of the ellipse, and in this case the ellipse is elongated in the y-direction.

Then c is the distance along the y-axis from the center to each focus:

\begin{align} c^2 &= a^2 - b^2 \\ &= 9 - 4 = 5 \end{align}

The distance is

$$c = \sqrt{5} \approx 2.2$$

Now we can sketch the ellipse. It helps to begin with the "box" into which the ellipse will be drawn or inscribed. That box has dimensions 2a × 2b, centered at the origin. Sketch in the vertices and the foci, and draw the ellipse, and you're done.

### Example 3

Sketch the ellipse (not centered at the origin) described by   $9(x - 1)^2 + 6(y + 2)^2 = 36$

Solution: Here we need to coax this equation

$$9(x - 1)^2 + 6(y + 2)^2 = 36$$

into one that looks like the standard form for an ellipse. This one is obviously not centered a the origin; we know that adding a number to x or y before squaring is a translation in the x- and y-direction, respectively.

$$\frac{9(x - 1)^2}{36} + \frac{6(y + 2)^2}{36} = 1$$

Dividing the 9 and 6 into the denominators (36) gives us the standard form, from which we can identify a and b:

$$a = \sqrt{6} \approx 2.45 \; \; b = 2$$

and because c2 = a2 - b2 for any ellipse, we have c:

$$c = \sqrt{2} \approx 1.41$$

All that remains is to sketch the graph. The dimensions of the bounding box (dashed line) are 2a × 2b.

The major axis is along the y-direction and the foci are located about 1.4 units above and below the center of the ellipse, at (1, -2).

The vertices are located at the center of each side of the bounding box, and the ellipse (our sketch, anyway) is a smooth curve connecting those. If you aren't familiar with the translation transformations we used here to move the ellipse over a unit and down two (or you've forgotten them, you might want to review horizontal translations on the functions page, or translations of an ellipse on the conic sections page.

### Example 4

Write an equation for an ellipse located in the x-y plane with its center at (-5, 6), and having a major axis of length 15 units in the x-direction, and a minor axis of length 10 units.

Solution: In a case like this, we can actually sketch our ellipse first: The center is at (-5, 6), as given. This means that we add 5 to x before squaring, and we subtract 6 from y in the y-term before squaring. These translations look like the opposite of what they are. If the length of the major axis is 15 units, then a = 15/2 = 7.5 units, and a2 = 56.25. Likewise, b = 6 and b2 = 36; we just plug those into our basic form and we have the equation of our ellipse.

### Tilted ellipses

We can apply one more transformation to an ellipse, and that is to rotate its axes by an angle, θ, about the center of the ellipse, or to tilt it. We can come up with a general equation for an ellipse tilted by θ by applying the 2-D rotational matrix to the vector (x, y) of coordinates of the ellipse. Here is the rotation matrix:

$$\left[ \begin{matrix} cos(\theta) && -sin(\theta) \\ sin(\theta) && cos(\theta) \end{matrix} \right]$$

Multiplication of (x,y) by the rotation matrix gives rotated vectors x and y.

$$\left[ \begin{matrix} cos(\theta) && -sin(\theta) \\ sin(\theta) && cos(\theta) \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] = \left[ \begin{matrix} x cos(\theta) - y sin(\theta) \\ x sin(\theta) + y cos(\theta) \end{matrix} \right]$$

Now if we recall the general form of an ellipse centered at (0, 0)

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

[Wide formulas on narrow screens: scroll formula L ↔ R]

we can insert our rotated x and y into it:

$$\frac{(x \cdot cos(\theta) - y \cdot sin(\theta))^2}{a^2} + \frac{(x \cdot sin(\theta) + y \cdot cos(\theta))^2}{b^2} = 1$$

Expanding the numerators gives:

$$\frac{x^2 cos^2(\theta) - 2xy \, sin(\theta) cos(\theta) + y^2 sin^2(\theta)}{a^2} + \frac{x^2 sin^2(\theta) + 2xy \, sin(\theta)cos(\theta) + y^2 cos^2(\theta)}{b^2} = 1$$

Now if we collect terms in x2, xy and y2 we get

$$\left( \frac{cos^2(\theta)}{a^2} + \frac{sin^2(\theta)}{b^2} \right)x^2 + 2 xy \, sin(\theta)cos(\theta) \left( \frac{1}{b^2} - \frac{1}{a^2} \right) + \left( \frac{sin^2(\theta)}{a^2} + \frac{cos^2(\theta)}{b^2} \right) y^2 = 1$$

This is a quadratic equation parameterized by the angle, θ, and characterized by constants A, B and C: f(θ) = Aθ2 + Bθ + C Notice that A and C are always positive because they consist of a sum of squares. The part of B in parentheses is positive because a > b for an ellipse, so the B term is negative only when the signs of sin(θ) and cos(θ) are opposite, that is, when the angle is in quadrant II and quadrant IV.  xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.