Ellipses

Learning goals

Define an ellipse by a constant sum of focal distances
Square twice to reach $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1$
Read $b^2 = a^2 - c^2$ as a right triangle
Find the major axis under the larger denominator
Shift the center by replacing $x$ with $x - h$
Measure eccentricity $e = \tfrac{c}{a}$ , zero for a circle

Two pins and a string

The circle in the previous lesson was built from one fixed point: every point of the circle is the same distance $r$ from the center. An ellipse is built from two fixed points, called the foci (singular: focus). A point is on the ellipse when its distances to the two foci add to a constant.

Wherever the pencil P sits on the curve, the two pieces of string add to the same total. Slide P around and r1 and r2 each change, but r1 + r2 never does.

The distance condition

Set up coordinates so the two foci sit symmetrically on the $x$ -axis, at $F_1(-c, 0)$ and $F_2(c, 0)$ with $c \ge 0$ . Call the constant total length of the string $2a$ . Writing the constant as $2a$ rather than as a single letter costs nothing now and saves a division by two later.

You need one tool to turn distances into algebra, and it is just the Pythagorean theorem. The segment from $(x_1, y_1)$ to $(x_2, y_2)$ is the hypotenuse of a right triangle whose legs are the horizontal gap $|x_2 - x_1|$ and the vertical gap $|y_2 - y_1|$ , so its length is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ . With that, the point $P(x, y)$ lies on the ellipse exactly when

\underbrace{\sqrt{(x + c)^2 + y^2}}_{r_1 \;=\; PF_1} \;+\; \underbrace{\sqrt{(x - c)^2 + y^2}}_{r_2 \;=\; PF_2} \;=\; 2a.

Before doing any algebra, notice that $a$ and $c$ cannot be just anything. Any trip from $F_1$ out to $P$ and back to $F_2$ is at least as long as the direct trip from $F_1$ to $F_2$ , which has length $2c$ . That is the triangle inequality, and it says $r_1 + r_2 \ge 2c$ , with equality only when $P$ sits on the segment between the foci. So there are three cases:

If $2a < 2c$ , the string is shorter than the gap between the pins and no point satisfies the condition.
If $2a = 2c$ , the string is pulled perfectly straight and the “curve” collapses to the segment joining the foci.
If $2a > 2c$ , there is slack, and you get a genuine closed curve.

So a real ellipse forces $a > c \ge 0$ . Keep that inequality in view: it is what makes the next step legal.

From the string to the equation

The distance condition is exactly $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ #

Start from the definition and isolate one radical, because two square roots cannot be cleared at once:

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

This is exactly the routine from Solving Radical Equations: get one radical alone, square, and repeat if a radical survives. Squaring both sides and using $(m - n)^2 = m^2 - 2mn + n^2$ with $m = 2a$ and $n = \sqrt{(x - c)^2 + y^2}$ , whose square is $(x - c)^2 + y^2$ , gives

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

Expand $(x + c)^2 = x^2 + 2cx + c^2$ on the left and $(x - c)^2 = x^2 - 2cx + c^2$ on the right. The $x^2$ , the $c^2$ , and the $y^2$ appear on both sides and cancel, which is the whole reason for putting the foci symmetrically about the origin. What survives is

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

Move the $-2cx$ across, move the radical term across, and divide by $4$ :

a\sqrt{(x - c)^2 + y^2} \;=\; a^2 - cx.

One radical is left and it is already alone, so square a second time. On the left, squaring $a$ and the radical gives $a^2\left[(x - c)^2 + y^2\right]$ ; on the right, $(a^2 - cx)^2 = a^4 - 2a^2cx + c^2x^2$ :

a^2\left(x^2 - 2cx + c^2 + y^2\right) \;=\; a^4 - 2a^2cx + c^2x^2.

The term $-2a^2cx$ appears on both sides and cancels. Gather the $x^2$ terms on the left and the constants on the right:

a^2x^2 - c^2x^2 + a^2y^2 \;=\; a^4 - a^2c^2.

Factor $a^2 - c^2$ out of the $x^2$ terms and out of the right-hand side:

(a^2 - c^2)\,x^2 + a^2y^2 \;=\; a^2(a^2 - c^2).

Here is where $a > c$ earns its keep. Because $a > c \ge 0$ , the number $a^2 - c^2$ is positive, so it is the square of a positive real number, and we may define $b > 0$ by

b^2 = a^2 - c^2, \qquad\text{equivalently}\qquad a^2 = b^2 + c^2.

Substituting $b^2$ turns the equation into $b^2x^2 + a^2y^2 = a^2b^2$ , and dividing by $a^2b^2$ , which is not zero, leaves the standard form:

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

One duty remains, and skipping it would be dishonest. Squaring is not reversible: from $u = v$ you may always conclude $u^2 = v^2$ , but from $u^2 = v^2$ you may only conclude $u = \pm v$ . That is precisely how radical equations manufacture extraneous solutions, and we squared twice. Each squaring is safe on the region where both sides are non-negative, and both sides are non-negative here. Even so, rather than track signs step by step, it is cleaner to check the converse outright.

Suppose $(x, y)$ satisfies $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ . Then $\frac{x^2}{a^2} \le 1$ , so $|x| \le a$ , and $y^2 = b^2\left(1 - \frac{x^2}{a^2}\right)$ . Substitute that $y^2$ into the squared distance from $P$ to $F_2$ , then use $1 - \frac{b^2}{a^2} = \frac{a^2 - b^2}{a^2} = \frac{c^2}{a^2}$ and $b^2 + c^2 = a^2$ :

\begin{aligned} (x - c)^2 + y^2 &= x^2\left(1 - \frac{b^2}{a^2}\right) - 2cx + \left(c^2 + b^2\right) \\ &= \frac{c^2x^2}{a^2} - 2cx + a^2 = \left(a - \frac{cx}{a}\right)^2. \end{aligned}

Since $|x| \le a$ and $0 \le c < a$ , the quantity $\frac{cx}{a}$ lies between $-c$ and $c$ , so $a - \frac{cx}{a}$ is at least $a - c > 0$ . It is positive, so taking the square root needs no absolute value: $r_2 = a - \frac{cx}{a}$ . Running the identical computation with $c$ replaced by $-c$ gives $r_1 = a + \frac{cx}{a}$ , also positive. Adding them,

r_1 + r_2 = \left(a + \frac{cx}{a}\right) + \left(a - \frac{cx}{a}\right) = 2a.

Every point of $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ really does lie on the string. The two squarings invented nothing and lost nothing, so the equation and the distance condition describe the same set of points.

That computation pays a bonus. It showed that the distance from a point of the ellipse to the focus $F_2$ is $r_2 = a - \frac{cx}{a}$ , and $x$ ranges over $[-a, a]$ , so $r_2$ ranges over $[\,a - c,\; a + c\,]$ . The closest a point of the ellipse ever gets to a focus is $a - c$ , and the farthest is $a + c$ . Both extremes happen at the vertices, the two ends of the long axis.

What $a$ , $b$ , and $c$ mean

Read the geometry straight off $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ .

Set $y = 0$ : then $x^2 = a^2$ , so $x = \pm a$ . The points $(\pm a, 0)$ are the vertices, and the segment joining them is the major axis, of length $2a$ . So the length of the string equals the length of the major axis, which is a fact worth remembering.

Set $x = 0$ : then $y^2 = b^2$ , so $y = \pm b$ . The points $(0, \pm b)$ are the co-vertices, and the segment joining them is the minor axis, of length $2b$ .

Since $b^2 = a^2 - c^2$ and $c \ge 0$ , we get $b \le a$ : the ellipse really is at least as wide as it is tall when the foci are on the $x$ -axis. And $a^2 = b^2 + c^2$ is not a formula to memorize, it is a right triangle you can point at. Go from the center straight up to a co-vertex $(0, b)$ , then across to a focus $(c, 0)$ . The two legs are $b$ and $c$ , so the hypotenuse has length $\sqrt{b^2 + c^2} = \sqrt{a^2} = a$ . The distance from a co-vertex to each focus is therefore exactly $a$ , and the two of them add to $2a$ , exactly as the string demands. The co-vertex is the one point where you can see the string length with no calculation at all.

The right triangle center, focus, co-vertex has legs c and b and hypotenuse a. That is the whole content of a squared equals b squared plus c squared, and it is why no side can beat the hypotenuse a.

Because $a$ is the hypotenuse of that triangle, no side can beat it: $a \ge b$ and $a > c$ , with $a = b$ precisely when $c = 0$ , which is the circle. So outside that one case $a$ is strictly the largest of the three, and that single observation kills most of the sign errors people make with ellipses.

Check your understanding

Where are the foci of $\frac{x^2}{169} + \frac{y^2}{144} = 1$ ?

(\pm 5, 0)

(0, \pm 5)

(\pm 13, 0)

(\pm \sqrt{313}, 0)

Which variable gets $a^2$

Nothing forced the foci onto the $x$ -axis. Put them at $(0, -c)$ and $(0, c)$ instead and run the same derivation with the roles of $x$ and $y$ interchanged. Every step is identical, so the result is identical with $x$ and $y$ swapped:

\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1, \qquad b^2 = a^2 - c^2, \qquad \text{foci } (0, \pm c).

So there is one rule to internalize, and it is the source of more wrong answers in this topic than anything else:

$a^2$ is always the larger denominator, and the major axis runs along the axis of the variable that sits above it. The foci lie on that same axis, $c$ units from the center, where $c^2 = (\text{larger denominator}) - (\text{smaller denominator})$ . (When the two denominators are equal there is no larger one: that is the circle, and it has no major axis.)

Notice you never have to remember which of $a$ and $b$ to subtract: $c$ is always found by subtracting the smaller denominator from the larger one. The reason is that $c$ is a leg of that right triangle and $a$ is the hypotenuse.

Worked example 1 Compare $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and $\frac{x^2}{9} + \frac{y^2}{25} = 1$

These two equations use the same numbers, and they are different ellipses.

For $\frac{x^2}{25} + \frac{y^2}{9} = 1$ , the larger denominator $25$ sits under $x^2$ , so $a^2 = 25$ and $b^2 = 9$ , giving $a = 5$ and $b = 3$ . The major axis is horizontal. Then

c^2 = a^2 - b^2 = 25 - 9 = 16 \quad\Rightarrow\quad c = 4.

The vertices are $(\pm 5, 0)$ , the co-vertices are $(0, \pm 3)$ , the foci are $(\pm 4, 0)$ , and the eccentricity (defined below) is $\frac{c}{a} = \frac{4}{5}$ . As a check, the distance from the co-vertex $(0, 3)$ to the focus $(4, 0)$ is $\sqrt{16 + 9} = 5 = a$ , so the two focal distances from that point add to $10 = 2a$ .

For $\frac{x^2}{9} + \frac{y^2}{25} = 1$ , the larger denominator $25$ now sits under $y^2$ , so $a^2 = 25$ still gives $a = 5$ , but the major axis is vertical. The value of $c$ is the same,

c^2 = 25 - 9 = 16 \quad\Rightarrow\quad c = 4,

but now everything is measured up and down: the vertices are $(0, \pm 5)$ , the co-vertices are $(\pm 3, 0)$ , and the foci are $(0, \pm 4)$ . The second ellipse is the first one turned on its side.

Check your understanding

Which axis is the major axis of $\frac{x^2}{4} + \frac{y^2}{49} = 1$ , and how long is it?

A Horizontal, length

4

B Vertical, length

7

C Vertical, length

14

D Horizontal, length

14

Moving the center off the origin

An ellipse does not have to be centered at the origin. From Transformations of Graphs you already know the move: replacing $x$ by $x - h$ shifts the graph right by $h$ , and replacing $y$ by $y - k$ shifts it up by $k$ . Doing both to the standard form slides the whole ellipse so its center lands at $(h, k)$ :

\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \qquad\text{(major axis horizontal)},

\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1 \qquad\text{(major axis vertical)}.

Shifting cannot change any length, so $a$ , $b$ , and $c$ mean exactly what they meant before. Everything is now measured from the center: the vertices are $a$ units from $(h, k)$ along the major axis. The co-vertices are $b$ units from the center along the minor axis, and the foci are $c$ units from the center along the major axis.

From a general equation to standard form

A general equation like $9x^2 + 4y^2 + 54x - 8y + 49 = 0$ hides an ellipse. To expose it, complete the square in $x$ and in $y$ , exactly as in the Completing the Square lesson, and then divide so the right side becomes $1$ .

Worked example 2 Put $9x^2 + 4y^2 + 54x - 8y + 49 = 0$ in standard form

Group the $x$ terms and the $y$ terms, and move the constant to the right side:

9x^2 + 54x + 4y^2 - 8y = -49.

Factor the leading coefficient out of each group, since completing the square needs a bare $x^2$ and a bare $y^2$ :

9\left(x^2 + 6x\right) + 4\left(y^2 - 2y\right) = -49.

Complete each square inside its parentheses. Half of $6$ is $3$ , and $3^2 = 9$ ; half of $-2$ is $-1$ , and $(-1)^2 = 1$ . Watch the bookkeeping: adding $9$ inside the first parentheses really adds $9 \cdot 9 = 81$ to the left side, and adding $1$ inside the second really adds $4 \cdot 1 = 4$ . So add those same amounts to the right side:

9\left(x^2 + 6x + 9\right) + 4\left(y^2 - 2y + 1\right) = -49 + 81 + 4 = 36.

Write each group as a square:

9(x + 3)^2 + 4(y - 1)^2 = 36.

Divide every term by $36$ so the right side is $1$ :

\frac{(x + 3)^2}{4} + \frac{(y - 1)^2}{9} = 1.

Now read it off. The center is $(-3, 1)$ , because $x + 3 = x - (-3)$ . The larger denominator, $9$ , sits under the $y$ term, so $a^2 = 9$ , $a = 3$ , and the major axis is vertical; $b^2 = 4$ , so $b = 2$ . Then

c^2 = a^2 - b^2 = 9 - 4 = 5 \quad\Rightarrow\quad c = \sqrt{5}.

Measuring from the center: the vertices are $(-3, 1 \pm 3)$ , that is $(-3, 4)$ and $(-3, -2)$ ; the co-vertices are $(-3 \pm 2, 1)$ , that is $(-1, 1)$ and $(-5, 1)$ ; and the foci are $(-3, 1 \pm \sqrt{5})$ .

Eccentricity, and the circle hiding inside every ellipse

Two ellipses can both have $a = 5$ and look nothing alike: one nearly round, one nearly flat. What distinguishes them is how far the foci sit from the center compared with the size of the curve. That ratio is the eccentricity:

e = \frac{c}{a}.

Since $0 \le c < a$ , the eccentricity always satisfies $0 \le e < 1$ . It also controls the shape completely, because $c = ea$ and therefore

b^2 = a^2 - c^2 = a^2 - e^2a^2 = a^2\left(1 - e^2\right), \qquad \frac{b}{a} = \sqrt{1 - e^2}.

At the flat end, pushing $e$ toward $1$ pushes $c$ toward $a$ , which squeezes $b$ toward $0$ and flattens the ellipse against its major axis. At the round end something better happens. Take $e = 0$ , which means $c = 0$ . The two foci merge into a single point at the center, $b^2 = a^2 - 0 = a^2$ gives $b = a$ , and the equation becomes

\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1 \quad\Longleftrightarrow\quad x^2 + y^2 = a^2,

a circle of radius $a$ . The definition degenerates just as gracefully: with $F_1 = F_2 = O$ , the condition $r_1 + r_2 = 2a$ becomes $2 \cdot PO = 2a$ , that is $PO = a$ , which is precisely “every point is distance $a$ from the center.” A circle is not a different kind of curve. It is the ellipse whose two foci have collided, and the circle lesson is the special case $e = 0$ of this one.

Three ellipses with the same major axis. Only the foci move. At e = 0 the foci sit together at the center and the curve is a circle; as e grows the foci separate and the curve flattens.

Eccentricity is how astronomers describe orbits. Earth’s orbit has $e \approx 0.017$ , so close to a circle that a drawing of it at page size is indistinguishable from one. Halley’s comet has $e \approx 0.97$ , a long thin cigar of an orbit. Both are ellipses with the sun at a focus.

Check your understanding

What curve is $\frac{x^2}{25} + \frac{y^2}{25} = 1$ ?

A An ellipse with foci

(\pm 5, 0)

B A circle of radius

5

C An ellipse with

c = 5

D A circle of radius

25

Building an ellipse from its geometry

Most problems hand you geometric facts (foci, a vertex, a string length, an eccentricity) and ask for the equation. The recipe never changes: find the center, decide which axis is major, then get any two of $a$ , $b$ , $c$ and use $a^2 = b^2 + c^2$ for the third.

Worked example 3 Foci $(2, -1)$ and $(2, 7)$ , with focal distances summing to $10$

The center is halfway between the foci, so average the coordinates:

\left(\frac{2 + 2}{2}, \; \frac{-1 + 7}{2}\right) = (2, 3).

Both foci have the same $x$ -coordinate, so they lie on the vertical line $x = 2$ . The foci always lie on the major axis, so the major axis is vertical, which tells you $a^2$ belongs under the $y$ term.

The distance between the foci is $2c$ , and here that distance is $7 - (-1) = 8$ , so $c = 4$ . The sum of the focal distances is $2a$ , and it is given as $10$ , so $a = 5$ . Check the requirement $a > c$ : indeed $5 > 4$ , so a genuine ellipse exists. Now find $b$ :

b^2 = a^2 - c^2 = 25 - 16 = 9.

Place the center at $(2, 3)$ with $a^2 = 25$ under the $y$ term:

\frac{(x - 2)^2}{9} + \frac{(y - 3)^2}{25} = 1.

Worked example 4 A semi-elliptical arch $30$ m wide and $10$ m high

A bridge arch is half an ellipse, cut along the major axis. Put the center of the full ellipse at the origin, with the road along the $x$ -axis. The arch spans $30$ meters, so the full major axis is $30$ and $a = 15$ . The arch is $10$ meters high at its middle, and that height is the semi-minor axis, so $b = 10$ . The upper half of

\frac{x^2}{225} + \frac{y^2}{100} = 1

is the arch. How high is it $9$ meters from the center? Substitute $x = 9$ :

\frac{81}{225} + \frac{y^2}{100} = 1 \quad\Rightarrow\quad \frac{y^2}{100} = 1 - \frac{9}{25} = \frac{16}{25}.

Multiply by $100$ and take the positive root, since the arch is the upper half:

y^2 = \frac{1600}{25} = 64 \quad\Rightarrow\quad y = 8.

The arch is $8$ meters high there. Note it is not $\frac{9}{15}$ of the way down from $10$ to $0$ : the curve is not a straight line, and only the equation gets it right.

Common mistakes

Calling the first denominator $a^2$ . In $\frac{x^2}{9} + \frac{y^2}{25} = 1$ the value of $a^2$ is $25$ , not $9$ . Always take $a^2$ to be the larger denominator, then let it tell you which axis is major. Getting this backwards makes every later answer wrong.
Adding when you should subtract. For an ellipse, $c^2 = a^2 - b^2$ . Computing $c^2 = a^2 + b^2$ is the single most common slip here. Look at the right triangle: $a$ is the hypotenuse, so no side beats it ( $a \ge b$ and $a > c$ ), and $c$ is found by taking away.
Putting the foci on the minor axis. The foci always lie on the major axis, inside the curve, at distance $c < a$ from the center. If your foci landed on the short axis, you swapped $a$ and $b$ .
Stopping before the right side is $1$ . After completing the square, $9(x + 3)^2 + 4(y - 1)^2 = 36$ is not standard form. Divide by $36$ first; only then are the denominators $a^2$ and $b^2$ .
Sign slips in the center. In $\frac{(x + 3)^2}{4} + \frac{(y - 1)^2}{9} = 1$ the center is $(-3, 1)$ , because $x + 3$ means $x - (-3)$ . Read the center as the value that makes each squared term zero.
Confusing $a$ with $2a$ . The string length, the sum of the focal distances, and the major axis are all $2a$ . The semi-major axis is $a$ . Halve or double at the right moment.
Thinking a circle is something else. A circle is the ellipse with $c = 0$ , $a = b$ , and $e = 0$ . It is not an exception to the definition, it is the middle of it.

Key takeaways

An ellipse with foci $F_1$ and $F_2$ is the set of points $P$ with $PF_1 + PF_2 = 2a$ , a constant. The triangle inequality forces $a > c$ , where $2c$ is the distance between the foci.
Isolating a radical and squaring twice turns that condition into the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ . The squarings lose nothing, because every point of the final equation is shown to satisfy the original distance condition.
$b^2 = a^2 - c^2$ , equivalently $a^2 = b^2 + c^2$ , is the right triangle with legs $b$ and $c$ and hypotenuse $a$ . So no side beats $a$ ( $a \ge b$ and $a > c$ , with $a = b$ exactly when $c = 0$ , the circle), and $c^2$ is always $(\text{larger denominator}) - (\text{smaller denominator})$ .
$a^2$ sits under the variable whose axis is the major axis, and the foci lie on that axis, $c$ from the center. Vertices are $a$ from the center, co-vertices are $b$ from the center.
A center at $(h, k)$ replaces $x$ by $x - h$ and $y$ by $y - k$ . Complete the square to get there from a general equation, then divide so the right side is $1$ .
The eccentricity $e = \frac{c}{a}$ satisfies $0 \le e < 1$ . It measures how far from circular the ellipse is: $e = 0$ is exactly a circle (the foci coincide), and $e$ near $1$ is nearly flat.
The distance from a point of the ellipse to a focus runs between $a - c$ and $a + c$ , and both extremes occur at the vertices.

Ellipses

Learning goals

Two pins and a string

The distance condition

From the string to the equation

What $a$ , $b$ , and $c$ mean

Which variable gets $a^2$

Moving the center off the origin

From a general equation to standard form

Eccentricity, and the circle hiding inside every ellipse

Building an ellipse from its geometry

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)

Ellipses

Learning goals

Two pins and a string

The distance condition

From the string to the equation

What aaa, bbb, and ccc mean

Which variable gets a2a^2a2

Moving the center off the origin

From a general equation to standard form

Eccentricity, and the circle hiding inside every ellipse

Building an ellipse from its geometry

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)

What $a$ , $b$ , and $c$ mean

Which variable gets $a^2$