🎯⭐ INTERACTIVE LESSON

Ellipses and Hyperbolas

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Ellipses and Hyperbolas - Complete Interactive Lesson

Part 1: Slicing the Cone

🛰️ Ellipses & Hyperbolas

Part 1 of 7 — Slicing the Cone

Topics in This Part

Section
What Is a Conic Section?
The Two Defining Distances
Anatomy of an Ellipse
Anatomy of a Hyperbola

🔑 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:

Slice angle	Curve
Perpendicular to the axis	Circle
Tilted, but cuts one nappe cleanly	Ellipse
Parallel to a slant edge	Parabola
Steep — cuts both nappes	Hyperbola

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, $2a$ :

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

Concept Check 🎯

Anatomy of an Ellipse

Picture a horizontal ellipse centered at the origin:

Feature	Meaning
Center	The midpoint of the ellipse, $(h, k)$
Major axis	The longer axis; length $2a$
Minor axis	The shorter axis; length $2b$

Anatomy of a Hyperbola

Now a horizontal hyperbola centered at the origin:

Feature	Meaning
Center	The midpoint between the two branches, $(h, k)$
Transverse axis	The axis that passes through both vertices; length $2a$
Vertices	The two "tips" where the branches turn around (distance $a$ from center)

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 ? A point lies on a hyperbola with . Its distances to the foci are and , where is on the branch nearer (so ). Find .

Part 2: The Standard Equations

🛰️ 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 $1$ :

$\frac{^{}}{}$

Part 3: Ellipses: Graphing & Finding Foci

🛰️ 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 , , then via . That hands you vertices, co-vertices, and foci — everything you need to graph.

Part 4: Hyperbolas: Asymptotes, Vertices & Foci

🛰️ 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 — the rails that guide the branches to infinity.

Part 5: Shifted Centers & Completing the Square

🛰️ 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 . When an equation is given expanded, to recover that standard form.

Part 6: Eccentricity & Real-World Applications

🛰️ 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: $e = \dfrac{c}{a}$

Part 7: Mixed Mastery & Exit Quiz

🛰️ 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

Question	Ellipse	Hyperbola
Sign between terms	$+$

💡 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.

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 $x^2$ .

⚠️ 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.

d(Q,F_1) > d(Q,F_2)

\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:

	Ellipse	Hyperbola
Sign between terms	$+$	$-$
Key relationship	$c^2 = a^2 - b^2$

⚠️ 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 term — whether that's the -term or the -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 the vertices, so and .

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 .

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 $\dfrac{x^2}{7} + \dfrac{y^2}{49} = 1$ : Hyperbola : Ellipse :

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: $a^{2}$

Step 2 — Find $c$ using the ellipse relationship $c^2 = a^2 - b^2$ :

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

Feature	Coordinates
Vertices	$(\pm 5,\, 0)$
Co-vertices	$(0,\, \pm 3)$
Foci	$(\pm 4,\, 0)$

✅ 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: $a^2 = 49 \Rightarrow a = 7, \qquad b^2 = 16 \Rightarrow b = 4$

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

Feature	Coordinates
Vertices	$(0,\, \pm 7)$
Co-vertices	$(\pm 4,\, 0)$
Foci	$(0,\, \pm\sqrt{33})$

⚠️ 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) $\dfrac{x^2}{100} + \dfrac{y^2}{64} = 1,\quad c = \,?$

Graphing Workflow 🔽

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

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 $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ :

💡 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 hyperbola, is under , so the slope is . 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 $c^2 = a^2 + b^2$ :

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

Feature	Coordinates / Equation
Vertices	$(\pm 3,\, 0)$
Foci	$(\pm 5,\, 0)$
Asymptotes	$y = \pm\dfrac{4}{3}x$

✅ 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) $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1,\quad c = \,?$

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 $= \,?$

complete the square

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 $(x + 3)^{2} = (x - (-$ puts the center at . The sign you see is the 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 $y^{2} - 2$ add — but it's inside a factor of , so you actually add to the right:

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 — 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.

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

Conic	Range of $e$	Meaning
Circle	$e = 0$	foci merged at center ( $c=0$ )
Ellipse	$0 < e < 1$	closer to $0$ = rounder; closer to $1$ = flatter
Parabola	$e = 1$	the boundary case
Hyperbola	$e > 1$	larger $e$ = more open branches

💡 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 🔽

⚠️ 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.

\dfrac{y^2}{8} - \dfrac{x^2}{40} = 1

\dfrac{x^2}{50} + \dfrac{y^2}{14} = 1

a^{2} = 25 \Rightarrow a = 5, b^{2} = 9 \Rightarrow b = 3

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

c^{2} = 25 - 9 = 16 \Rightarrow c = 4

c^2 = a^2 - b^2 = 49 - 16 = 33 \Rightarrow c = \sqrt{33}\approx 5.7

\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1,\quad c = \,?

\dfrac{x^2}{20} + \dfrac{y^2}{4} = 1,\quad c = \,?

y = \pm\frac{a}{b}\,x

c^2 = 9 + 16 = 25 \Rightarrow c = 5

c^{2} = 9 + 16 = 25 \Rightarrow c = 5

\dfrac{y^2}{64} - \dfrac{x^2}{36} = 1,\quad c = \,?

\dfrac{x^2}{1} - \dfrac{y^2}{8} = 1,\quad c = \,?

(x + 3)^{2} = (x - (- 3))^{2}

h)(horizontal hyperbola)

(x^2 + 4x + 4) + 4(y^2 - 2y + 1) = 8 + 4 + 4

(x+2)^2 + 4(y-1)^2 = 16

Ellipses and Hyperbolas

Ellipses and Hyperbolas - Complete Interactive Lesson

Part 1: Slicing the Cone

🛰️ Ellipses & Hyperbolas

Topics in This Part

What Is a Conic Section?

The Two Defining Distances

Anatomy of an Ellipse

Anatomy of a Hyperbola

Part 2: The Standard Equations

🛰️ Ellipses & Hyperbolas

The Two Standard Forms (centered at the origin)

Part 3: Ellipses: Graphing & Finding Foci

🛰️ Ellipses & Hyperbolas

Part 4: Hyperbolas: Asymptotes, Vertices & Foci

🛰️ Ellipses & Hyperbolas

Part 5: Shifted Centers & Completing the Square

🛰️ Ellipses & Hyperbolas

Part 6: Eccentricity & Real-World Applications

🛰️ Ellipses & Hyperbolas

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

Part 7: Mixed Mastery & Exit Quiz

🛰️ Ellipses & Hyperbolas

Quick Reference Card

The aaa, bbb, ccc Relationships — They Differ!

Orientation: Which Way Does It Open?

Worked Example: x225+y29=1\dfrac{x^2}{25} + \dfrac{y^2}{9} = 125x2​+9y2​=1

Worked Example: x216+y249=1\dfrac{x^2}{16} + \dfrac{y^2}{49} = 116x2​+49y2​=1

Asymptotes: The Defining Feature

Worked Example: x29−y216=1\dfrac{x^2}{9} - \dfrac{y^2}{16} = 19x2​−16y2​=1

Standard Form with Center (h,k)(h, k)(h,k)

Completing the Square to Find Standard Form

Where These Curves Show Up

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

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

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

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

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

Standard Form with Center $(h, k)$