The ellipse

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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.

Anatomy of an ellipse: terminology

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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

Drawing an ellipse

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Geometry of an ellipse

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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.

The derivation

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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:

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

Example 1

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

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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:

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

Example 2

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

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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

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

Example 3

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

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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.

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.

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Solution: In a case like this, we can actually sketch our ellipse first:

Tilted ellipses

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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.

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[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.

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