Ellipses
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 7.5: “Conic Sections” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/7-5-conic-sections |
| Textbook used in class | Stewart, Calculus, Section 10.5: “Conic Sections” (Examples 1, 2) |
Opening Scenario
The orbit of every planet, moon, and satellite is an ellipse. So is the shape of a whispering gallery, where a sound made at one focus reflects off the curved wall and arrives clearly at the other focus. Both phenomena follow from the defining property of the ellipse: every point on it is exactly the same total distance from two fixed points.
Quick Reference
Definition. An ellipse is the set of all points $P$ in the plane such that the sum of distances from $P$ to two fixed points $F_1$ and $F_2$ (the foci) equals a constant $2a$: \[ |PF_1| + |PF_2| = 2a. \]
Standard forms with center at the origin:
| Foci on the | Equation | Vertices on | Semi-axes |
|---|---|---|---|
| $x$-axis | $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1,\quad a > b > 0$ | $(\pm a, 0)$ and $(0, \pm b)$ | major $a$, minor $b$ |
| $y$-axis | $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1,\quad a > b > 0$ | $(0, \pm a)$ and $(\pm b, 0)$ | major $a$, minor $b$ |
Key relationship: \[ c^2 = a^2 - b^2, \quad \text{so } c = \sqrt{a^2 - b^2}, \] where $c$ is the distance from the center to each focus. Because $a > b$, we have $c < a$.
Eccentricity: $e = \dfrac{c}{a}$, and $0 < e < 1$ for every ellipse. A circle is the degenerate case $e = 0$ ($c = 0$, both foci at the center).
Key Concepts
1. Why the Sum-of-Distances Definition Produces an Oval
Place two thumbtacks in a board, tie a string of length $2a$ between them, and trace the curve keeping the string taut against a pencil. The pencil moves so that its distance to one tack plus its distance to the other tack always equals the string length $2a$. The resulting closed curve is an ellipse. The tacks are the foci; the string length is $2a$.
When both foci coincide at the same point, the two distances are equal and their sum is the diameter, so the curve is a circle. Pulling the foci apart while keeping the string length fixed stretches the circle into a flatter oval.
2. Reading the Parameters From the Equation
Given an equation in standard form, the larger denominator is always $a^2$, and the foci lie on the axis corresponding to that larger denominator.
Example 1. Find the foci, vertices, and eccentricity of $9x^2 + 16y^2 = 144$. (This is Stewart 10.5, Example 1.)
Goal. Rewrite in standard form by dividing through by $144$, then read off $a$, $b$, $c$.
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1. \]
Here $a^2 = 16 > b^2 = 9$, and the larger denominator is under $x^2$, so the major axis lies along the $x$-axis. \[ a = 4, \quad b = 3, \quad c = \sqrt{16 - 9} = \sqrt{7}. \]
Boxed answer:
- Foci: $(\pm\sqrt{7},\, 0)$.
- Vertices on the major axis: $(\pm 4,\, 0)$.
- Vertices on the minor axis: $(0,\, \pm 3)$.
- Eccentricity: $e = \dfrac{\sqrt{7}}{4} \approx 0.66$.
Recap. Dividing first is the essential step. Once the right side is $1$, the parameters read off directly from the denominators. The foci are closer to the center than the vertices ($\sqrt{7} < 4$), which must always be true since $c < a$.
3. Finding the Equation From the Geometric Data
The information given is usually the location of the foci and vertices. Use those to find $a$, $b$, and $c$, then substitute into the appropriate standard form.
Example 2. Find an equation of the ellipse with foci $(0,\, \pm 2)$ and vertices $(0,\, \pm\sqrt{13})$. (This is Stewart 10.5, Example 2.)
Goal. The foci are on the $y$-axis, so use the second standard form with $a^2$ under $y^2$.
The foci give $c = 2$ and the vertices give $a = \sqrt{13}$. Then \[ b^2 = a^2 - c^2 = 13 - 4 = 9. \]
The standard form with major axis along the $y$-axis is $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$, so:
\[ \frac{x^2}{9} + \frac{y^2}{13} = 1. \]
Boxed answer: $\dfrac{x^2}{9} + \dfrac{y^2}{13} = 1$.
Recap. The foci determine which axis is major. Once you know $a$ and $c$, compute $b^2 = a^2 - c^2$. Notice $b^2 = 9 > 0$, which confirms the data are geometrically consistent (you need $a > c$ for an ellipse to exist).
4. Eccentricity and Shape
Eccentricity $e = c/a$ is a single number that captures how elongated the ellipse is.
- When $e \approx 0$, the foci are near the center and the ellipse is nearly circular.
- When $e \approx 1$, the foci are near the vertices and the ellipse is very flat.
Earth’s orbit has $e \approx 0.017$, nearly circular. Some comets have $e$ close to $1$, making their orbits very elongated.
confusing $a^2 - b^2$ with $a^2 + b^2$. The foci of an ellipse lie inside the oval, so $c < a$, which forces $c^2 = a^2 - b^2$ (not a sum). Students who memorize $c^2 = a^2 + b^2$ are thinking of the hyperbola formula. For an ellipse, the minus sign is correct because the foci are interior to the curve.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using $c^2 = a^2 + b^2$ for an ellipse | $c = \sqrt{16 + 9} = 5$ | Ellipse: $c^2 = a^2 - b^2$. The $+$ formula belongs to the hyperbola |
| Taking the smaller denominator as $a^2$ | calling $9$ the value of $a^2$ in $\frac{x^2}{16} + \frac{y^2}{9} = 1$ | $a^2$ is always the larger denominator; $a = 4$, not $3$ |
| Placing foci on the wrong axis | foci at $(0, \pm c)$ when larger denominator is under $x^2$ | Foci are on the axis of the larger denominator |
| Forgetting to divide the equation first | reading $a^2 = 144$ from $9x^2 + 16y^2 = 144$ | Divide by $144$ first so the right side is $1$ |
Common Misconceptions
$c^2 = a^2 + b^2$ for an ellipse.
This is the concept-image-conflicts-definition error. Students who have just studied the Pythagorean theorem or who confuse the ellipse formula with the hyperbola formula write a plus sign instead of a minus sign. For an ellipse the foci lie inside the oval, so $c < a$, which forces $c^2 = a^2 - b^2$. A counterexample: for $\frac{x^2}{16} + \frac{y^2}{9} = 1$, using $c^2 = 16 + 9 = 25$ gives $c = 5$, but the foci would then lie outside the ellipse (since $c = 5 > a = 4$), which contradicts the definition.
Leveled Practice
Level 1 -- Direct Reading
Problem 1. Write the equation $25x^2 + 4y^2 = 100$ in standard form and identify $a$, $b$, $c$, and the foci.
Show answer
Divide by $100$: $\dfrac{x^2}{4} + \dfrac{y^2}{25} = 1$.
Here $a^2 = 25$ is under $y^2$, so the major axis is the $y$-axis. $a = 5$, $b = 2$, $c = \sqrt{25 - 4} = \sqrt{21}$.
Boxed answer: Foci at $(0,\, \pm\sqrt{21})$.
Problem 2. Find the equation of the ellipse with foci $(\pm 3, 0)$ and one vertex at $(5, 0)$.
Show answer
Foci on the $x$-axis with $c = 3$ and $a = 5$, so $b^2 = 25 - 9 = 16$.
Boxed answer: $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.
Level 2 -- Multi-Step
Problem 3. An ellipse has foci $(\pm\sqrt{5}, 0)$ and passes through the point $(2, 1)$. Find its equation.
Show answer
The foci are on the $x$-axis, so the equation has the form $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with $a^2 > b^2$ and $c = \sqrt{5}$, giving $b^2 = a^2 - 5$.
Substituting the point $(2, 1)$: \[ \frac{4}{a^2} + \frac{1}{a^2 - 5} = 1. \] Multiply through by $a^2(a^2 - 5)$: \[ 4(a^2 - 5) + a^2 = a^2(a^2 - 5). \] \[ 5a^2 - 20 = a^4 - 5a^2. \] \[ a^4 - 10a^2 + 20 = 0. \] Wait -- let me redo: $4(a^2-5) + a^2 = a^2(a^2-5)$, so $5a^2 - 20 = a^4 - 5a^2$, giving $a^4 - 10a^2 + 20 = 0$. Let $u = a^2$: $u = (10 \pm \sqrt{100 - 80})/2 = (10 \pm \sqrt{20})/2 = 5 \pm \sqrt{5}$.
Since $a^2 > c^2 = 5$, take $a^2 = 5 + \sqrt{5}$ and $b^2 = \sqrt{5}$.
Boxed answer: $\dfrac{x^2}{5+\sqrt{5}} + \dfrac{y^2}{\sqrt{5}} = 1$.
(This problem illustrates that not all ellipses have clean integer parameters.)
Level 3 -- Conceptual
Problem 4. Explain why the eccentricity of an ellipse must satisfy $0 < e < 1$, using the relationship $c^2 = a^2 - b^2$ and the requirement $a > b > 0$.
Show answer
From $c^2 = a^2 - b^2$ with $b > 0$, we get $c^2 < a^2$, so $c < a$, meaning $e = c/a < 1$.
From $c^2 = a^2 - b^2 \geq 0$ (since $a^2 \geq b^2$), we get $c \geq 0$. Equality holds when $b = a$, which gives a circle ($e = 0$). A true ellipse has $b < a$, so $c > 0$ and $e > 0$.
Therefore $0 < e < 1$ for every proper ellipse.
Mastery Checklist
Mental Model
Think of $a$, $b$, and $c$ as the sides of a right triangle whose hypotenuse is $a$. The relationship $c^2 = a^2 - b^2$ is just the Pythagorean theorem rearranged: $a^2 = b^2 + c^2$. Draw the triangle: one leg is $b$ (half the minor axis), the other is $c$ (the focal distance), and the hypotenuse is $a$ (half the major axis). This triangle, called the focal triangle, sits with its right angle at the end of the minor axis and its hypotenuse running from the center to a vertex of the major axis. Keeping that triangle in mind makes $c^2 = a^2 - b^2$ impossible to confuse with the hyperbola formula.
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