Ellipses and Hyperbolas

🛰️ Ellipses & Hyperbolas

Part 1 of 7 — Slicing the Cone

Topics in This Part

🔑 Key Concept: An ellipse and a hyperbola both come from slicing a double cone. Each is defined by a rule about distances to two fixed points called foci. The ellipse adds those distances; the hyperbola subtracts them. Master that one contrast and everything else follows.

What Is a Conic Section?

A conic section is the curve you get when a flat plane slices through a double cone. The angle of the slice decides the shape:

In this lesson we focus on the two "closed-vs-open" twins:

• An ellipse is a closed, oval loop (a circle is the special case where it is perfectly round).
• A hyperbola is two separate open branches that fly apart forever.

💡 A circle is just an ellipse whose two foci have merged into a single center. Everything you learn about ellipses contains circles as a special case.

The Two Defining Distances

Every ellipse and hyperbola has two special interior points called foci (singular: focus). The whole curve is built from a distance rule involving those foci.

Ellipse — the sum is constant. For every point $P$ on an ellipse, the sum of the distances to the two foci is the same fixed number, $2a$:

$d(P, F_1) + d(P, F_2) = 2a$

💡 The "thumbtack and string" trick: pin two tacks (the foci), loop a string around them, and trace with a pencil pulled taut. Because the string length never changes, the sum of distances stays constant — and you draw a perfect ellipse.

Hyperbola — the difference is constant. For every point $P$ on a hyperbola, the absolute difference of the distances to the two foci is fixed at $2a$:

$\left| d(P, F_1) - d(P, F_2) \right| = 2a$

🔑 One word — sum vs. difference — is the entire difference between the two curves' definitions.

Concept Check 🎯

Anatomy of an Ellipse

Picture a horizontal ellipse centered at the origin:

Because the major axis is the longer one, we always have $a > b > 0$ for an ellipse. The foci live inside the ellipse on the major axis.

⚠️ For an ellipse, $a$ is always the bigger of $a$ and $b$ — $a$ is whichever number sits under the longer axis, not necessarily under $x^2$.

Anatomy of a Hyperbola

Now a horizontal hyperbola centered at the origin:

⚠️ For a hyperbola, $a$ is not necessarily larger than $b$. Here $a$ is whatever sits under the positive term, and the foci are even farther out than the vertices, so $c > a$. We'll nail the $a, b, c$ relationships in Part 2 — they are different for each curve.

Match the Vocabulary 🔽

Choose the correct term for each description.

Apply the Definitions 🧮

Use the distance rules directly. (No equations needed — just the focal-distance definitions.)

1) A point $P$ lies on an ellipse with $2a = 20$. Its distance to one focus is $d(P, F_1) = 7$. What is $d(P, F_2)$? 2) A point $Q$ lies on a hyperbola with $2a = 12$. Its distances to the foci are $d(Q, F_1) = 15$ and $d(Q, F_2) = ?$, where $Q$ is on the branch nearer $F_2$ (so $d(Q,F_1) > d(Q,F_2)$). Find $d(Q, F_2)$.

🛰️ Ellipses & Hyperbolas

Part 2 of 7 — The Standard Equations

🔑 The Goal: Read a standard-form equation and instantly extract the center, the orientation, and the values of $a$, $b$, and $c$. The plus/minus sign between the two terms is the single most important clue.

The Two Standard Forms (centered at the origin)

Ellipse — the two terms are added and the right side is $1$:

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

Hyperbola — the two terms are subtracted:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \qquad\text{or}\qquad \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

🔑 The sign tells you the shape. A plus sign means ellipse; a minus sign means hyperbola. Always rearrange to "$= 1$" before reading anything off.

The $a$, $b$, $c$ Relationships — They Differ!

This is the table students most often confuse. The relationship between $a$, $b$, and $c$ is opposite for the two curves:

⚠️ Do not memorize "$a$ is under $x^2$." For an ellipse, $a^2$ is the larger denominator (under the longer axis). For a hyperbola, $a^2$ is the denominator of the positive term — whether that's the $x$-term or the $y$-term.

Why the sign flip in the formula? In an ellipse the foci sit inside (closer to center than the vertices), so $c < a$ and $c^2 = a^2 - b^2$. In a hyperbola the foci sit outside the vertices, so $c > a$ and $c^2 = a^2 + b^2$.

Concept Check 🎯

Orientation: Which Way Does It Open?

Ellipse. The longer axis lies along the variable with the larger denominator.

• Larger denominator under $x^2$ → major axis is horizontal (wider than tall).
• Larger denominator under $y^2$ → major axis is vertical (taller than wide).

Hyperbola. The branches open along the variable of the positive term.

• $x^2$ positive → opens left/right (horizontal transverse axis).
• $y^2$ positive → opens up/down (vertical transverse axis).

💡 For a hyperbola, the size of the denominators does not decide orientation — only the sign does. $\dfrac{y^2}{4} - \dfrac{x^2}{25} = 1$ opens up/down even though $25 > 4$.

Read the Orientation 🔽

For each equation, choose the conic type and how it opens.

Pick Out $a^2$ 🧮

Remember: for an ellipse $a^2$ is the larger denominator; for a hyperbola $a^2$ is the denominator of the positive term. Enter the value of $a^2$.

1) Ellipse $\dfrac{x^2}{7} + \dfrac{y^2}{49} = 1$: $a^2 = \,?$ 2) Hyperbola $\dfrac{y^2}{8} - \dfrac{x^2}{40} = 1$: $a^2 = \,?$ 3) Ellipse $\dfrac{x^2}{50} + \dfrac{y^2}{14} = 1$: $a^2 = \,?$

🛰️ Ellipses & Hyperbolas

Part 3 of 7 — Ellipses: Graphing & Finding Foci

🔑 The Workflow: From $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, find $a$, $b$, then $c$ via $c^2 = a^2 - b^2$. That hands you vertices, co-vertices, and foci — everything you need to graph.

Worked Example: $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$

Step 1 — Identify $a^2$ and $b^2$. The larger denominator is under $x^2$, so the major axis is horizontal: $a^2 = 25 \Rightarrow a = 5, \qquad b^2 = 9 \Rightarrow b = 3$

Step 2 — Find $c$ using the ellipse relationship $c^2 = a^2 - b^2$: $c^2 = 25 - 9 = 16 \Rightarrow c = 4$

Step 3 — List the key points (center at origin, horizontal major axis):

✅ Sanity check: the foci must lie inside the ellipse on the major axis. $c = 4 < a = 5$ ✓ — foci are between the center and the vertices, exactly as they should be.

Worked Example: $\dfrac{x^2}{16} + \dfrac{y^2}{49} = 1$

Now the larger denominator ($49$) is under $y^2$, so the major axis is vertical: $a^2 = 49 \Rightarrow a = 7, \qquad b^2 = 16 \Rightarrow b = 4$ $c^2 = a^2 - b^2 = 49 - 16 = 33 \Rightarrow c = \sqrt{33}\approx 5.7$

Because the major axis is vertical, the vertices and foci are on the $y$-axis:

⚠️ The foci always sit on the major axis (here the $y$-axis), never on the minor axis. If your foci came out on the short axis, you mixed up $a$ and $b$.

Concept Check 🎯

Find $c$ 🧮

Compute the focal distance $c$ for each ellipse using $c^2 = a^2 - b^2$. (Round to two decimals if not a whole number.)

1) $\dfrac{x^2}{100} + \dfrac{y^2}{64} = 1,\quad c = \,?$ 2) $\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1,\quad c = \,?$ 3) $\dfrac{x^2}{20} + \dfrac{y^2}{4} = 1,\quad c = \,?$

Graphing Workflow 🔽

You're graphing $\dfrac{x^2}{36} + \dfrac{y^2}{100} = 1$. Fill in each step.

🛰️ Ellipses & Hyperbolas

Part 4 of 7 — Hyperbolas: Asymptotes, Vertices & Foci

🔑 The Workflow: From standard form, the positive term gives $a$ and the orientation; $c^2 = a^2 + b^2$ gives the foci; and the denominators give the asymptotes — the rails that guide the branches to infinity.

Asymptotes: The Defining Feature

For a hyperbola centered at the origin, the asymptotes pass through the center with slopes built from $a$ and $b$:

Horizontal $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$: $y = \pm\frac{b}{a}\,x$

Vertical $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$: $y = \pm\frac{a}{b}\,x$

💡 The box trick. Mark $\pm a$ along the transverse axis and $\pm b$ along the other axis, draw the rectangle through those four points, and extend its diagonals — those diagonals are exactly the asymptotes. The vertices touch the box at the midpoints of two sides.

⚠️ The slope is (rise of the $y$-related length) over (run of the $x$-related length). For a horizontal hyperbola, $a$ is under $x^2$, so the slope is $b/a$. For a vertical hyperbola, $a$ is under $y^2$, so the slope is $a/b$. When in doubt, rebuild the box.

Worked Example: $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

Step 1 — Orientation & $a$, $b$. The $x^2$ term is positive, so it opens left/right: $a^2 = 9 \Rightarrow a = 3, \qquad b^2 = 16 \Rightarrow b = 4$

Step 2 — Find $c$ using the hyperbola relationship $c^2 = a^2 + b^2$: $c^2 = 9 + 16 = 25 \Rightarrow c = 5$

Step 3 — Asymptotes (horizontal, slope $\pm b/a$): $y = \pm\frac{4}{3}\,x$

✅ Sanity check: for a hyperbola the foci are farther out than the vertices, so $c > a$. Here $5 > 3$ ✓.

Concept Check 🎯

Find $c$ for Hyperbolas 🧮

Use $c^2 = a^2 + b^2$ (note the PLUS for hyperbolas). Round to two decimals if not whole.

1) $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1,\quad c = \,?$ 2) $\dfrac{y^2}{64} - \dfrac{x^2}{36} = 1,\quad c = \,?$ 3) $\dfrac{x^2}{1} - \dfrac{y^2}{8} = 1,\quad c = \,?$

Asymptote Slopes 🧮

Enter the POSITIVE asymptote slope as a fraction or decimal. (The other asymptote is its negative.)

1) $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1$ → slope $= \,?$ 2) $\dfrac{y^2}{9} - \dfrac{x^2}{4} = 1$ → slope $= \,?$

🛰️ Ellipses & Hyperbolas

Part 5 of 7 — Shifted Centers & Completing the Square

🔑 The Big Idea: Move the center from the origin to $(h, k)$ by replacing $x$ with $(x-h)$ and $y$ with $(y-k)$. When an equation is given expanded, complete the square to recover that standard form.

Standard Form with Center $(h, k)$

Just shift each variable. The center moves, but $a$, $b$, $c$ and the shapes are unchanged:

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

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

⚠️ Watch the signs. $(x-h)^2$ means the center's $x$-coordinate is $+h$. So $(x+3)^2 = (x-(-3))^2$ puts the center at $x=-3$. The sign you see is the opposite of the coordinate.

To get vertices/foci, find them as if centered at the origin, then add $(h, k)$ to shift them into place. Asymptotes of a shifted hyperbola pass through $(h,k)$: $y - k = \pm\frac{b}{a}(x - h) \quad \text{(horizontal hyperbola)}$

Concept Check 🎯

Completing the Square to Find Standard Form

Example: $\;x^2 + 4x + 4y^2 - 8y - 8 = 0$.

Step 1 — Group and factor out leading coefficients (the $+$ between groups already hints "ellipse"): $(x^2 + 4x) + 4(y^2 - 2y) = 8$

Step 2 — Complete each square. For $x^2+4x$ add $(4/2)^2=4$. For $y^2-2y$ add $(2/2)^2=1$ — but it's inside a factor of $4$, so you actually add $4\cdot 1 = 4$ to the right: $(x^2 + 4x + 4) + 4(y^2 - 2y + 1) = 8 + 4 + 4$ $(x+2)^2 + 4(y-1)^2 = 16$

Step 3 — Divide by $16$ to get the "$=1$" form: $\frac{(x+2)^2}{16} + \frac{(y-1)^2}{4} = 1$

✅ Center $(-2, 1)$, $a^2=16$, $b^2=4$. Since the $+$ remained, it's an ellipse — as the matching signs predicted.

Complete the Square 🧮

The hyperbola $9x^2 - 4y^2 - 36x - 8y + 68 = 0$ rewrites as $\frac{(y - k)^2}{9} - \frac{(x - h)^2}{4} = 1.$ Find the center coordinates.

1) $h = \,?$ 2) $k = \,?$

Shift It 🔽

A hyperbola has standard form $\dfrac{(x-3)^2}{4} - \dfrac{(y+1)^2}{9} = 1$. Fill in each feature.

🛰️ Ellipses & Hyperbolas

Part 6 of 7 — Eccentricity & Real-World Applications

🔑 Eccentricity $e = \dfrac{c}{a}$ measures how "stretched" a conic is. It is the single number that distinguishes a nearly-circular orbit from a wildly elongated one, and it's how astronomers and engineers describe these curves.

Eccentricity: $e = \dfrac{c}{a}$

Eccentricity always equals (focal distance) ÷ (semi-major or semi-transverse length):

💡 For an ellipse, $c < a$ so $e < 1$ automatically. For a hyperbola, $c > a$ so $e > 1$ automatically. The value of $e$ tells you the conic type at a glance.

Example (ellipse). $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ had $a=5$, $c=4$, so $e = \dfrac{4}{5} = 0.8$ — a fairly flattened ellipse.

Example (hyperbola). $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ had $a=3$, $c=5$, so $e = \dfrac{5}{3} \approx 1.67$.

Concept Check 🎯

Compute Eccentricity 🧮

Find $e = c/a$. Round to two decimals.

1) Ellipse $\dfrac{x^2}{100}+\dfrac{y^2}{36}=1$ (so $a=10$, $c=8$): $e = \,?$ 2) Hyperbola $\dfrac{x^2}{9}-\dfrac{y^2}{27}=1$ (so $a=3$, $c=6$): $e = \,?$

Where These Curves Show Up

Ellipses

• Planetary orbits. Kepler's First Law: planets orbit the Sun in ellipses with the Sun at one focus (not the center).
• Whispering galleries. Sound from one focus reflects off an elliptical ceiling and converges precisely at the other focus — whisper at one spot, heard clearly across the room.
• Lithotripsy. Shock waves emitted from one focus of an elliptical reflector focus on a kidney stone placed at the other focus.

Hyperbolas

• Navigation (LORAN/GPS-style). A constant difference in signal arrival times from two stations places you on a hyperbola — exactly its defining property.
• Cooling towers & some telescope mirrors use hyperbolic cross-sections.
• Comet paths. A comet with $e > 1$ swings past the Sun once on a hyperbolic trajectory and never returns.

🔑 The reflective "focus-to-focus" property of the ellipse and the "constant difference" property of the hyperbola are not trivia — they are exactly the definitions from Part 1 doing real work.

Match the Application 🔽

🛰️ Ellipses & Hyperbolas

Part 7 of 7 — Mixed Mastery & Exit Quiz

You can now (1) tell the curves apart by sign, (2) extract $a$, $b$, $c$ correctly for each, (3) graph with vertices, foci, and asymptotes, (4) handle shifted centers, and (5) compute eccentricity. Time to put it all together.

Quick Reference Card

⚠️ Top traps: (1) using $-$ instead of $+$ in the focal relation (or vice-versa); (2) flipping the asymptote slope between $b/a$ and $a/b$; (3) forgetting the sign flip when reading a shifted center.

Mixed Practice 🎯

Mixed Computation 🧮

1) For the ellipse $\dfrac{x^2}{49}+\dfrac{y^2}{24}=1$, find $c$. 2) For the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$, find the eccentricity $e$ (two decimals). 3) The hyperbola $\dfrac{x^2}{36}-\dfrac{y^2}{64}=1$ has positive asymptote slope $= \,?$ (fraction ok).

Full Analysis 🔽

Analyze $\dfrac{(x-1)^2}{16} - \dfrac{(y+2)^2}{9} = 1$ completely.

Exit Quiz ✅

Answer all three to finish the lesson.