Definition via focus, directrix, eccentricity e: Worked examples

You have seen the parent note build the rule $\frac{P F}{P M} = e$ . Now we drill it until no case can surprise you. Every sign of the directrix, every value of $e$ , the degenerate limits, a word problem, and an exam twist — each gets its own fully worked example.

The scenario matrix

Every cell below is hit by at least one worked example on this page.

#	Case class	Concrete knob setting	Example
A	Directrix left of focus, $e = 1$	$x = - a$ , parabola	Ex 1
B	Directrix right of focus, $0 < e < 1$	$x = + d$ , ellipse	Ex 2
C	Directrix right of focus, $e > 1$	$x = + d$ , hyperbola	Ex 3
D	Horizontal directrix	$y = + d$ , opens down	Ex 4
E	Degenerate: $e \to 0$	circle limit	Ex 5
F	Word problem (real world)	satellite dish, $e = 1$	Ex 6
G	Exam twist: recover $e, F, L$ from an equation	reverse-engineer	Ex 7
H	Edge case: focus lies close, extract latus rectum	semi-latus rectum $ℓ$	Ex 8

Prerequisites you may want open: Distance formula and distance from a point to a line and Latus rectum and semi-latus rectum.

Ex 1 — Cell A: directrix on the LEFT, $e = 1$ (parabola)

Worked example Statement

Focus $F = (2, 0)$ , directrix $x = - 2$ , eccentricity $e = 1$ . Find the curve.

Forecast: Equal distances to a point and a vertical line — guess the shape and where it opens before reading on.

Step 1. $P F = (x - 2)^{2} + y^{2}$ . Why this step? Straight distance formula from $P = (x, y)$ to the point $(2, 0)$ .

Step 2. $P M = ∣ x - (- 2) ∣ = ∣ x + 2∣$ . Why this step? Distance to the vertical line $x = - 2$ is purely horizontal — the $y$ -coordinate is irrelevant. Look at the green dashed segment in the figure: it is flat.

Step 3. Set $P F = e \cdot P M = 1 \cdot ∣ x + 2∣$ and square: $(x - 2)^{2} + y^{2} = (x + 2)^{2} .$ Why square? Both sides are non-negative distances, so squaring is safe and it removes the root and the absolute value at once.

Step 4. Expand and cancel: $x^{2} - 4 x + 4 + y^{2} = x^{2} + 4 x + 4 \Rightarrow y^{2} = 8 x .$ Why did the $x^{2}$ vanish? Because $e = 1$ : the two squared brackets are identical except for the sign of the middle term, so $x^{2}$ subtracts out. This is the algebraic fingerprint of a parabola.

Verify: Compare $y^{2} = 8 x$ with the standard $y^{2} = 4 a x$ (see Parabola - Standard equation y^2=4ax). Then $4 a = 8 \Rightarrow a = 2$ , so the focus should be at $(a, 0) = (2, 0)$ ✅ and the directrix $x = - a = - 2$ ✅ — exactly our inputs.

Figure — Definition via focus, directrix, eccentricity e

Ex 2 — Cell B: directrix on the RIGHT, $0 < e < 1$ (ellipse)

Worked example Statement

Focus $F = (0, 0)$ , directrix $x = 8$ , eccentricity $e = \frac{1}{2}$ . Find the curve.

Forecast: Since $e < 1$ the point must stay closer to the focus, so the curve should close up. Ellipse — but where is its centre?

Step 1. $P F = x^{2} + y^{2}$ , $P M = ∣ x - 8∣$ . Why the $∣ x - 8∣$ ? The directrix is now to the right, at $x = 8$ , so the horizontal gap is $x - 8$ (magnitude taken).

Step 2. $P F = \frac{1}{2} P M$ : $x^{2} + y^{2} = \frac{1}{2} ∣ x - 8∣$ .

Step 3. Square: $x^{2} + y^{2} = \frac{1}{4} (x - 8)^{2}$ . Why square? Same reason — non-negative sides, kill root and modulus.

Step 4. Multiply by $4$ : $4 x^{2} + 4 y^{2} = x^{2} - 16 x + 64$ . Why multiply? Clears the fraction so we can collect integer coefficients.

Step 5. Collect: $3 x^{2} + 16 x + 4 y^{2} = 64$ . Why bounded? Both $x^{2}$ and $y^{2}$ carry positive coefficients ( $3$ and $4$ ) → the curve cannot run off to infinity → ellipse ✅.

Step 6 (optional polish). Complete the square in $x$ : $3 (x + \frac{8}{3})^{2} + 4 y^{2} = 64 + \frac{64}{3} = \frac{256}{3} .$ Why complete the square? It exposes the centre at $(- \frac{8}{3}, 0)$ , matching the form in Ellipse - Standard form and c^2 = a^2 - b^2.

Verify: Take the point $x = y = 0$ (the focus). Then $P F = 0$ and $P M = 8$ , ratio $0/8 = 0 \neq = \frac{1}{2}$ , so the focus is not on the curve (correct — it's inside). Now test a genuine point: put $y = 0$ in $3 x^{2} + 16 x = 64$ , i.e. $3 x^{2} + 16 x - 64 = 0 \Rightarrow x = \frac{- 16 \pm 256 + 768}{6} = \frac{- 16 \pm 32}{6}$ , giving vertices $x = \frac{8}{3}$ and $x = - 8$ . Check $x = \frac{8}{3}$ : $P F = \frac{8}{3}$ , $P M = ∣ \frac{8}{3} - 8∣ = \frac{16}{3}$ , ratio $= \frac{8/3}{16/3} = \frac{1}{2} = e$ ✅.

Ex 3 — Cell C: directrix on the RIGHT, $e > 1$ (hyperbola)

Worked example Statement

Focus $F = (0, 0)$ , directrix $x = 1$ , eccentricity $e = 2$ . Find the curve.

Forecast: $e > 1$ lets a point sit farther from the focus than from the line — the curve should open into two branches.

Step 1. $x^{2} + y^{2} = 2∣ x - 1∣$ . Why $2$ ? $e = 2$ multiplies the perpendicular distance.

Step 2. Square: $x^{2} + y^{2} = 4 (x - 1)^{2} = 4 x^{2} - 8 x + 4$ .

Step 3. Move everything to one side: $0 = 3 x^{2} - 8 x + 4 - y^{2} \Rightarrow 3 x^{2} - y^{2} - 8 x + 4 = 0.$ Why one side? To read the signs. Here $x^{2}$ has coefficient $+ 3$ while $y^{2}$ has $- 1$ — opposite signs → hyperbola ✅ (compare Hyperbola - Standard form and c^2 = a^2 + b^2).

Step 4. Why two branches? Solve for $y^{2} = 3 x^{2} - 8 x + 4$ . For large $x$ this is positive and grows without bound, so $y \to \pm \infty$ — the curve escapes to infinity on both sides, never closing.

Verify: Find where $y = 0$ : $3 x^{2} - 8 x + 4 = 0 \Rightarrow x = \frac{8 \pm 64 - 48}{6} = \frac{8 \pm 4}{6}$ , giving vertices $x = 2$ and $x = \frac{2}{3}$ . Test $x = \frac{2}{3}, y = 0$ : $P F = \frac{2}{3}$ , $P M = ∣ \frac{2}{3} - 1∣ = \frac{1}{3}$ , ratio $= 2 = e$ ✅.

Ex 4 — Cell D: HORIZONTAL directrix (opens downward)

Worked example Statement

Focus $F = (0, 3)$ , directrix $y = - 3$ , eccentricity $e = 1$ . Find the curve.

Forecast: The directrix is a horizontal line now, and the focus is above it. A $e = 1$ curve equidistant from both should be a parabola opening upward.

Step 1. $P F = x^{2} + (y - 3)^{2}$ . Why? Distance to the point $(0, 3)$ .

Step 2. $P M = ∣ y - (- 3) ∣ = ∣ y + 3∣$ . Why $∣ y + 3∣$ and not $∣ x \pm d ∣$ ? The directrix is horizontal, so the perpendicular gap is vertical — it depends on $y$ , not $x$ . This is the twin of the earlier rule; look at the figure's vertical green segment.

Step 3. $P F = P M$ , square: $x^{2} + (y - 3)^{2} = (y + 3)^{2}$ .

Step 4. Expand: $x^{2} + y^{2} - 6 y + 9 = y^{2} + 6 y + 9 \Rightarrow x^{2} = 12 y$ . Why did $y^{2}$ cancel? Same $e = 1$ fingerprint — but this time the $x$ -role and $y$ -role are swapped, so it's the $y^{2}$ that disappears, leaving $x^{2} = 12 y$ : a parabola opening upward.

Verify: Standard form $x^{2} = 4 a y$ gives $4 a = 12 \Rightarrow a = 3$ , so focus $(0, a) = (0, 3)$ ✅ and directrix $y = - a = - 3$ ✅.

Ex 5 — Cell E: the DEGENERATE limit $e \to 0$ (circle)

Worked example Statement

Take $F = (0, 0)$ and directrix $x = d$ with $e$ tiny. What happens as $e \to 0$ if we push the directrix to infinity so that $e d$ stays fixed?

Forecast: $e = 0$ should collapse the curve to a circle centred at the focus (see Circle as limiting case e=0). But $P M \to \infty$ as the wall recedes — how does the equation survive?

Step 1. General squared equation with directrix $x = d$ : $x^{2} + y^{2} = e^{2} (x - d)^{2} = e^{2} x^{2} - 2 e^{2} d x + e^{2} d^{2} .$ Why start general? We want to watch the coefficients as $e \to 0$ .

Step 2. Let $e \to 0$ with the product $k = e d$ held fixed (push the wall out as $e$ shrinks). Why hold $e d$ fixed? If we just set $e = 0$ with $d$ finite the right side vanishes and we'd get only the single point $(0, 0)$ — a doubly degenerate limit. Holding $e d = k$ keeps a finite-size circle.

Step 3. As $e \to 0$ : $e^{2} x^{2} \to 0$ and $2 e^{2} d x = 2 e (e d) x = 2 e k x \to 0$ , while $e^{2} d^{2} = (e d)^{2} = k^{2}$ stays. We are left with $x^{2} + y^{2} = k^{2} .$ Why this is a circle: $x^{2} + y^{2} = r^{2}$ is the equation of a circle of radius $r = k$ centred at the origin (the focus). The directrix has vanished to infinity and stopped mattering.

Verify: Every point of $x^{2} + y^{2} = k^{2}$ is exactly distance $k$ from the origin, so $P F = k$ is constant — the hallmark of a circle. With $k = 5$ , the point $(3, 4)$ satisfies $3^{2} + 4^{2} = 25 = 5^{2}$ ✅ and $P F = 9 + 16 = 5$ ✅.

Ex 6 — Cell F: WORD PROBLEM (satellite dish)

Worked example Statement

A cross-section of a satellite dish is a parabola. The receiver (focus) sits $0.5 m$ above the vertex, and the dish "wall" (directrix) sits $0.5 m$ below the vertex. Place the vertex at the origin. Find the equation, and how wide the dish is at the height of the receiver.

Forecast: Focus above the vertex, directrix below → a parabola opening upward (just like Ex 4). The width at focus-height is a famous quantity — take a guess at its name.

Step 1. Vertex at origin, focus $(0, 0.5)$ , directrix $y = - 0.5$ , and $e = 1$ (all parabolas). Why $e = 1$ ? A dish is a parabola; parabolas are the $e = 1$ conic.

Step 2. Equidistance: $x^{2} + (y - 0.5)^{2} = ∣ y + 0.5∣$ , square and simplify exactly as in Ex 4: $x^{2} = 2 y .$ Why the coefficient $2$ ? Standard $x^{2} = 4 a y$ with $a = 0.5$ gives $4 a = 2$ .

Step 3. Width at receiver height means $y = a = 0.5$ (the focus level). Substitute: $x^{2} = 2 (0.5) = 1 \Rightarrow x = \pm 1.$ Why plug $y = 0.5$ ? We want the horizontal span of the curve exactly where the receiver sits.

Step 4. The full width is $2 \times 1 = 2 m$ . Why is this special? This chord through the focus, perpendicular to the axis, is the latus rectum; here it equals $4 a = 2 m$ (see Latus rectum and semi-latus rectum).

Verify: Latus rectum of $x^{2} = 4 a y$ is $4 a = 4 (0.5) = 2 m$ ✅, and units are metres throughout ✅.

Ex 7 — Cell G: EXAM TWIST (reverse-engineer $e$ , focus, directrix)

Worked example Statement

You are handed the conic $16 x^{2} + 25 y^{2} = 400$ and told it came from a focus–directrix rule. Recover $e$ , a focus, and the matching directrix.

Forecast: Two positive squared coefficients → ellipse, so we expect $0 < e < 1$ . Which axis is longer?

Step 1. Divide by $400$ : $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ . Why divide? To reach standard ellipse form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ so we can read $a$ and $b$ .

Step 2. Read $a^{2} = 25, b^{2} = 16$ , so $a = 5, b = 4$ (major axis along $x$ since $25 > 16$ ).

Step 3. $c^{2} = a^{2} - b^{2} = 25 - 16 = 9 \Rightarrow c = 3$ . Why this formula? For an ellipse the focal distance obeys $c^{2} = a^{2} - b^{2}$ (from Ellipse - Standard form and c^2 = a^2 - b^2).

Step 4. Eccentricity $e = \frac{c}{a} = \frac{3}{5} = 0.6$ . Why $c / a$ ? This is the geometric formula for $e$ established in the parent note; it agrees with the focus–directrix ratio.

Step 5. A focus is at $(c, 0) = (3, 0)$ . The directrix is the vertical line $x = \frac{a}{e} = \frac{5}{3/5} = \frac{25}{3}$ . Why $x = a / e$ ? For a standard ellipse the directrix corresponding to focus $(+ c, 0)$ sits at $x = a / e$ (equivalently $a^{2} / c = 25/3$ ).

Verify: Check the vertex $(5, 0)$ with focus $(3, 0)$ and directrix $x = 25/3$ : $P F = ∣5 - 3∣ = 2$ , $P M = ∣5 - 25/3∣ = ∣ - 10/3 ∣ = 10/3$ , ratio $= \frac{2}{10/3} = \frac{6}{10} = 0.6 = e$ ✅. Also $a^{2} / c = 25/3 = 8. \overline{3}$ ✅.

Ex 8 — Cell H: EDGE CASE, extract the semi-latus rectum $ℓ$

Worked example Statement

For the parabola of Ex 1, $y^{2} = 8 x$ (focus $(2, 0)$ , $e = 1$ ), find the semi-latus rectum $ℓ$ — the distance from the focus straight up to the curve.

Forecast: "Semi" means half the latus-rectum chord. Guess: is it $2 a$ , $a$ , or something else?

Step 1. The latus rectum is the vertical chord through the focus. Set $x = 2$ (the focus's $x$ ) in $y^{2} = 8 x$ : $y^{2} = 16 \Rightarrow y = \pm 4.$ Why $x = 2$ ? The chord passes through the focus at $x = 2$ ; we intersect it with the curve.

Step 2. Full chord length $= 4 - (- 4) = 8$ . The semi-latus rectum is half: $ℓ = 4$ . Why half? "Semi" = from the focus to one side only, i.e. $y = + 4$ distance from $(2, 0)$ up to $(2, 4)$ , which is $4$ .

Step 3. Cross-check with the formula $ℓ = 2 a$ . Here $4 a = 8 \Rightarrow a = 2$ , so $2 a = 4 = ℓ$ ✅. Why $ℓ = 2 a$ ? For $y^{2} = 4 a x$ the latus rectum is $4 a$ , so its half is $2 a$ (see Latus rectum and semi-latus rectum).

Verify: The point $(2, 4)$ lies on $y^{2} = 8 x$ : $4^{2} = 16 = 8 (2)$ ✅, and its distance from the focus $(2, 0)$ is $4 = ℓ$ ✅.

Recall Scenario checklist — can you place any new problem?

Given a fresh problem, answer these and you already know the shape and method: Is the directrix vertical or horizontal? ::: Vertical → use $∣ x \pm d ∣$ ; horizontal → use $∣ y \pm d ∣$ . Which side is the directrix on? ::: Sets the sign inside the modulus; nothing else changes. What does $e$ tell you before any algebra? ::: $e = 0$ circle, $0 < e < 1$ ellipse, $e = 1$ parabola, $e > 1$ hyperbola. After squaring, how do you name the conic from coefficients? ::: One squared term missing → parabola; same signs → ellipse; opposite signs → hyperbola. How do you sanity-check any answer? ::: Pick a vertex, compute $P F / P M$ , confirm it equals $e$ .

Recall check

Parabola from

F = (2, 0)

, directrix

x = - 2

e = 1

y^{2} = 8 x

, with

a = 2

Ellipse from

F = (0, 0)

, directrix

x = 8

e = \frac{1}{2}

3 x^{2} + 16 x + 4 y^{2} = 64

Hyperbola from

F = (0, 0)

, directrix

x = 1

e = 2

3 x^{2} - y^{2} - 8 x + 4 = 0

Parabola from

F = (0, 3)

, directrix

y = - 3

e = 1

x^{2} = 12 y

, opening upward,

a = 3

Limit of the focus–directrix curve as

e \to 0

(with

e d

fixed

= k

The circle

x^{2} + y^{2} = k^{2}

For

16 x^{2} + 25 y^{2} = 400

, find

e

and a focus.

e = 3/5 = 0.6

, focus

(3, 0)

, directrix

x = 25/3

Semi-latus rectum of

y^{2} = 8 x

ℓ = 4

(equals

2 a

with

a = 2

How to check a focus–directrix answer at a vertex?

Compute

P F / P M

there; it must equal

e

Connections

3.4.01 Definition via focus, directrix, eccentricity e (Hinglish)
Parabola - Standard equation y^2=4ax
Ellipse - Standard form and c^2 = a^2 - b^2
Hyperbola - Standard form and c^2 = a^2 + b^2
Circle as limiting case e=0
Distance formula and distance from a point to a line
Latus rectum and semi-latus rectum

Worked examples — Definition via focus, directrix, eccentricity e

The scenario matrix

Ex 1 — Cell A: directrix on the LEFT, $e = 1$ (parabola)

Ex 2 — Cell B: directrix on the RIGHT, $0 < e < 1$ (ellipse)

Ex 3 — Cell C: directrix on the RIGHT, $e > 1$ (hyperbola)

Ex 4 — Cell D: HORIZONTAL directrix (opens downward)

Ex 5 — Cell E: the DEGENERATE limit $e \to 0$ (circle)

Ex 6 — Cell F: WORD PROBLEM (satellite dish)

Ex 7 — Cell G: EXAM TWIST (reverse-engineer $e$ , focus, directrix)

Ex 8 — Cell H: EDGE CASE, extract the semi-latus rectum $ℓ$

Recall check

Connections

Concept Map