Worked examples — Ellipse — standard forms, semi-major - minor axes, foci, eccentricity, latus rectum

Before anything, some tiny reminders (so no symbol is used un-earned):

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The scenario matrix

Here is every distinct kind of ellipse question. Each row is a "cell" — an example below is tagged with the cell it covers.

Cell	Case class	What makes it different	Example
A	Horizontal major, already standard	bigger denominator under $x^2$	Ex 1
B	Vertical major, already standard	bigger denominator under $y^2$	Ex 2
C	Not yet "= 1" (needs normalising)	RHS is not $1$	Ex 3
D	Build equation from data (foci + $2a$)	given geometry, find equation	Ex 4
E	Degenerate: $a=b$ ⇒ circle ($e=0$)	foci merge, minor case	Ex 5
F	Limiting: $e\to 1$ (very flat)	nails race to the ends	Ex 6
G	Real-world word problem	translate English → $a,b,c$	Ex 7 (whispering gallery)
H	Exam twist: find equation from $e$ + latus rectum	reverse-engineer $a,b$	Ex 8
I	Off-centre ellipse, centre $(h,k)$	curve shifted away from origin	Ex 9

Cells cover: both orientations, both degenerate/limiting ends of $e$, non-standard input, forward and reverse problems, an applied one, and a shifted (off-origin) ellipse. Nothing is left unseen.

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Example 1 — Cell A (horizontal major, standard)

Forecast: Guess first — which axis is the long one, $x$ or $y$?

The figure below shows this exact ellipse: the yellow segment along the $x$-axis is the major axis (from vertex to vertex), the pink vertical segment is the minor axis, the two blue dots are the foci at $(\pm 4,0)$, and the dashed pink line is the latus rectum standing on the right-hand focus.

1. Identify the bigger denominator. Here $25>9$, and $25$ sits under $x^2$. Why this step? The larger denominator always marks the major axis; that decides everything else's orientation.
2. Read off $a$ and $b$. $a^2=25\Rightarrow a=5$; $b^2=9\Rightarrow b=3$. Why? $a$ is the bigger half-length by definition, so it must be $\sqrt{25}$.
3. Compute $c$. $c^2=a^2-b^2=25-9=16\Rightarrow c=4$. Why? This is the derived relation $b^2=a^2-c^2$ rearranged; it places the foci.
4. Place the points. Vertices $(\pm 5,0)$, co-vertices $(0,\pm 3)$, foci $(\pm 4,0)$ — all on the $x$-axis because the major axis is horizontal (look at the blue nails in the figure).
5. Eccentricity. $e=c/a=4/5=0.8$. Why? $e$ measures squash; $0.8$ is fairly squashed but still an ellipse.
6. Latus rectum. $\text{LR}=\frac{2b^2}{a}=\frac{2\cdot 9}{5}=\frac{18}{5}=3.6$. Why? It is the vertical chord through a focus — a quick size gauge (the dashed pink line in the figure).

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Example 2 — Cell B (vertical major, standard)

Forecast: Same numbers as Ex 1, just swapped. Where do the foci go now?

The figure shows the same-sized ellipse but standing upright: now the yellow major axis runs vertically along the $y$-axis, and the blue foci dots have moved onto the $y$-axis at $(0,\pm 4)$. Compare it side by side with the Ex 1 figure to see the flip.

1. Bigger denominator? $25$, now under $y^2$. Why? This flips the major axis to vertical — the classic trap the parent warned about.
2. $a$ and $b$. $a^2=25\Rightarrow a=5$ (still the bigger!), $b^2=9\Rightarrow b=3$. Why? $a$ is always the larger, regardless of which letter it sits under.
3. $c$. $c^2=25-9=16\Rightarrow c=4$. Same as before. Why? The bridge $c^2=a^2-b^2$ depends only on the values $a$ and $b$, not on their orientation, so the arithmetic is identical to Ex 1.
4. Foci location. Because the major axis is vertical, foci sit on the $y$-axis: $(0,\pm 4)$. Why? Foci always live on the major axis (the nails must be on the long line).
5. Eccentricity. $e=4/5=0.8$. Why? $e=c/a$ is a ratio of lengths; rotating the ellipse changes neither $c$ nor $a$, so $e$ is unchanged from Ex 1.

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Example 3 — Cell C (needs normalising to "= 1")

Forecast: This isn't "$=1$" yet. What single operation fixes that?

1. Divide through by 36. $\frac{9x^2}{36}+\frac{4y^2}{36}=1\Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1$. Why? The standard form requires the right side to be exactly $1$; only then do denominators equal $a^2$ and $b^2$.
2. Bigger denominator. $9$ under $y^2$ ⇒ vertical major axis. Why? The larger denominator marks the major axis, exactly as in Ex 1–2; here it points at $y$.
3. $a,b$. $a^2=9\Rightarrow a=3$; $b^2=4\Rightarrow b=2$. Why? $a$ is the bigger half-length, so it must be the square root of the bigger denominator ($9$).
4. $c$. $c^2=9-4=5\Rightarrow c=\sqrt{5}\approx 2.236$. Foci $(0,\pm \sqrt{5})$. Why? The relation $c^2=a^2-b^2$ places the foci; they sit on the $y$-axis because that's the major axis.
5. $e$. $e=\frac{\sqrt{5}}{3}\approx 0.745$. Why? By definition $e=c/a$; this measures how squashed the ellipse is.
6. LR. $\frac{2b^2}{a}=\frac{2\cdot 4}{3}=\frac{8}{3}\approx 2.667$. Why? LR $=2b^2/a$ is the width through a focus — our quick size gauge.

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Example 4 — Cell D (build equation from foci + string length)

Forecast: Which of $a,b,c$ do the given facts hand you directly?

1. Read $a$ from the string. $2a=10\Rightarrow a=5$. Why? The constant sum $P{F}_1+P{F}_2$ (defined at the top: the two nail-to-point distances added) equals $2a$ (the loop of string length) — that's the definition.
2. Read $c$ from the foci. Foci at $(\pm 3,0)$ ⇒ $c=3$, and they're on the $x$-axis ⇒ horizontal major. Why? $c$ is by definition the distance from centre to a focus; here that distance is $3$.
3. Get $b$. $b^2=a^2-c^2=25-9=16$. Why? The Pythagorean relation (Pythagorean theorem) is the only bridge from $a,c$ to $b$.
4. Assemble. $\frac{x^2}{25}+\frac{y^2}{16}=1$. Why? Plug $a^2=25$ under $x^2$ (horizontal major) and $b^2=16$ under $y^2$.
5. Eccentricity. $e=c/a=3/5=0.6$. Why? By definition; $0.6$ confirms a moderately squashed ellipse.

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Example 5 — Cell E (degenerate: $a=b$ gives a circle, $e=0$)

Forecast: Both denominators are equal. Is this even an ellipse in the usual sense?

The figure shows the result: a perfect circle of radius $4$, with a single blue dot at the centre where both foci have collapsed together, and a yellow radius segment marking the constant distance $4$ from centre to curve.

1. Compare denominators. $16=16$ — there is no "bigger" one. Why this matters? When $a=b$, no axis is longer; the "major/minor" distinction dissolves.
2. Compute $c$. $c^2=a^2-b^2=16-16=0\Rightarrow c=0$. Why? With $c=0$, both foci sit at the same place: the centre $(0,0)$.
3. Eccentricity. $e=c/a=0/4=0$. Why? $e=0$ is the boundary of the range $0\le e<1$ — the least squashed possible.
4. Name it. With $a=b$ and $e=0$, this is a circle of radius $4$: $x^2+y^2=16$. (See Circle.) Why does $a=b$, $e=0$ force a circle? Equal half-axes mean every direction from the centre reaches the curve at the same distance $a=b=4$; "same distance in all directions from a point" is exactly the definition of a circle, and its single (merged) focus is that centre.

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Example 6 — Cell F (limiting: very flat, $e\to 1$)

Forecast: Guess — as $b$ gets tiny, does $e$ rise toward $0$ or toward $1$?

The figure overlays both ellipses on the same axes: the blue one ($b=4$) is a gentle oval with its foci well inside, while the pink one ($b=1$) is a thin sliver whose foci (pink dots) have crept out almost to the vertices. Watch how flattening the curve drives the foci toward the ends.

1. Case (i): $b=4$. $c=\sqrt{25-16}=\sqrt{9}=3$, so $e=3/5=0.6$.
2. Case (ii): $b=1$. $c=\sqrt{25-1}=\sqrt{24}\approx 4.899$, so $e=\sqrt{24}/5\approx 0.980$. Why? As $b\to 0$, $c=\sqrt{a^2-b^2}\to a$, so $e=c/a\to 1$.
3. Interpret. Smaller $b$ ⇒ foci race outward toward the vertices ⇒ ellipse flattens into a sliver. Why $e$ never reaches $1$? Because $b>0$ always keeps $c<a$ strictly; $e=1$ would need $b=0$, which is a line segment (a Parabola sits exactly at $e=1$ as the next conic).

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Example 7 — Cell G (real-world word problem: whispering gallery)

Forecast: "Width" and "height" — which becomes $2a$ and which $2b$?

1. Translate the words. Full width $=100$ m along the floor ⇒ major axis $2a=100\Rightarrow a=50$. Full height $=40$ m ⇒ minor axis $2b=40\Rightarrow b=20$. Why? The wider span is the longest way across = major axis; the shorter span = minor axis.
2. Find $c$ (focus distance from centre). $c^2=a^2-b^2=2500-400=2100\Rightarrow c=\sqrt{2100}\approx 45.826$ m.
3. Distance between the two people. They stand at $(\pm c,0)$, so separation $=2c=2\sqrt{2100}\approx 91.652$ m. Why? The foci are the two "listening spots" of an ellipse — sound from one reflects to the other.
4. Ceiling height above a focus. That's half the latus rectum: $\frac{b^2}{a}=\frac{400}{50}=8$ m. Why? Setting $x=c$ gave $y=\pm b^2/a$; the ceiling above the focus is the top value $b^2/a$.

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Example 8 — Cell H (exam twist: from eccentricity + latus rectum)

Forecast: You're given ratios, not lengths. Can two equations pin down two unknowns?

1. Write the two given facts as equations. $e=\frac{c}{a}=\frac{3}{5}\Rightarrow c=\frac{3a}{5},\text{LR}=\frac{2b^2}{a}=\frac{32}{5}.$ Why? Each given datum is a formula in $a,b,c$; two facts + the relation $b^2=a^2-c^2$ give three equations for three unknowns.
2. Express $b^2$ via $e$. $b^2=a^2-c^2=a^2-\frac{9a^2}{25}=\frac{16a^2}{25}$. Why? Substitute $c=\frac{3a}{5}$ to get $b^2$ purely in terms of $a$.
3. Feed into the LR equation. $\frac{2b^2}{a}=\frac{2}{a}\cdot \frac{16a^2}{25}=\frac{32a}{25}=\frac{32}{5}.$
4. Solve for $a$. From $\frac{32a}{25}=\frac{32}{5}$, multiply both sides by $\frac{25}{32}$: $a=\frac{32}{5}\cdot \frac{25}{32}=\frac{25}{5}=5$. Why this step? We now have a single equation in the single unknown $a$ (both $b$ and $c$ were rewritten in terms of $a$), so plain algebra isolates $a$ — the $32$'s cancel and the $25/5$ leaves $5$.
5. Back-substitute. $a=5\Rightarrow a^2=25$; $b^2=\frac{16\cdot 25}{25}=16$.
6. Equation. $\frac{x^2}{25}+\frac{y^2}{16}=1$.

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Example 9 — Cell I (off-centre ellipse, centre $(h,k)$)

Forecast: This looks like Ex 1 but with $(x-2)$ and $(y+1)$. Guess — does the shape change, or just its position?

The figure shows the same $5\times 3$ oval as Ex 1, but slid so its centre sits at the blue dot $(2,-1)$ instead of the origin. Every named point simply travels along with the centre.

1. Read the centre. The pattern $(x-h{)}^2$ and $(y-k{)}^2$ means the centre is $(h,k)$. Here $x-2$ gives $h=2$, and $y+1=y-(-1)$ gives $k=-1$. Centre $=(2,-1)$. Why? Replacing $x$ by $x-h$ shifts the whole curve right by $h$; the standard form is just the origin case with new "local coordinates" $X=x-h$, $Y=y-k$.
2. Read $a,b$ exactly as before. Bigger denominator $25$ under $(x-2{)}^2$ ⇒ horizontal major, $a^2=25\Rightarrow a=5$, $b^2=9\Rightarrow b=3$. Why? The denominators still play the identical role; the shift never touches them, only the position.
3. Find $c$. $c^2=a^2-b^2=25-9=16\Rightarrow c=4$. Why? Same bridge $c^2=a^2-b^2$; shifting the ellipse cannot change internal distances.
4. Place points around the centre $(2,-1)$. Move $\pm a$ horizontally and $\pm c$ horizontally from the centre:
   • Vertices: $(2\pm 5,-1)=(7,-1)$ and $(-3,-1)$.
   • Foci: $(2\pm 4,-1)=(6,-1)$ and $(-2,-1)$. Why? In local coordinates the vertices are $(\pm 5,0)$ and foci $(\pm 4,0)$; converting back with $x=X+2$, $y=Y-1$ just adds the centre offset.
5. Eccentricity. $e=c/a=4/5=0.8$ — unchanged from Ex 1. Why? Sliding an ellipse leaves all its lengths (hence the ratio $c/a$) exactly as they were.

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Recall One-line summary of the whole matrix

Every ellipse question reduces to finding two of $\{a,b,c,e\}$ and using $b^2=a^2-c^2$ (with $e=c/a$, $\text{LR}=2b^2/a$) to get the rest — then checking which axis is long by asking "where is the bigger denominator / the wider span / the foci?". For an off-centre curve, do all of this in local coordinates, then add the centre $(h,k)$ back.

Connections

• 3.4.04 Ellipse — standard forms, semi-major - minor axes, foci, eccentricity, latus rectum (index 3.4.4) (parent — the machinery these examples exercise)
• Circle (Ex 5: the $e=0$ degenerate case)
• Parabola (Ex 6: the $e=1$ boundary the ellipse never reaches)
• Pythagorean theorem (the $a^2=b^2+c^2$ triangle behind every $c$-computation)
• Eccentricity and directrix (deeper meaning of the $e$ we compute)
• Hyperbola — standard forms (contrast: uses $c^2=a^2+b^2$ and difference of focal distances)
• Conic Sections — overview (where all these cases sit together)