Definition via focus, directrix, eccentricity e: Exercises

Figure 1 below shows how the single dial $e$ slides you between an ellipse, a parabola, and a hyperbola (all sharing one focus and one vertical directrix). Figure 2 shows the two distances you will compute in almost every problem: $P F$ (to the focus point) and $P M$ (perpendicular, to the point $M$ on the directrix). Glance back at Figure 2 whenever a solution writes down $P F$ and $P M$ .

Figure 1 — one dial $e$ changes the whole shape. Caption / alt-text: on a warm cream grid with focus $F$ marked at the origin and the vertical dashed directrix $x = - d$ to its left, three nested curves are drawn from the same focus and directrix — a teal ellipse ( $e = 0.5$ , a closed oval), a burnt-orange parabola ( $e = 1$ , an open U), and a plum hyperbola ( $e = 1.8$ , two branches flying apart) — illustrating that raising $e$ opens the curve more and more.

Figure — Definition via focus, directrix, eccentricity e

Figure 2 — the two distances $P F$ and $P M$ , and the projection point $M$ . Caption / alt-text: on the same warm cream grid, the focus $F = (0, 0)$ and a sample point $P = (x, y)$ are marked; a burnt-orange arrow labelled " $P F$ (to point)" runs slanted from $F$ to $P$ , while a deep-teal horizontal arrow labelled " $P M = ∣ x + d ∣$ (perpendicular)" runs from $P$ straight across to the point $M$ on the dashed vertical directrix $x = - d$ , meeting it at a small right-angle square — showing $P F$ is a point-distance but $P M$ is measured perpendicular to the line.

Level 1 — Recognition

Goal: read $e$ off a given fact and name the curve. No algebra yet.

Q1.1 A conic has eccentricity $e = 0.5$ . Name the curve.

Q1.2 A conic satisfies $P F = P M$ for every point $P$ (distance to focus equals perpendicular distance to directrix). What is $e$ , and what curve is it?

Q1.3 For a certain conic, points are allowed to sit farther from the focus than from the directrix. What can you say about $e$ , and which curve is this?

Recall Solution — L1

Q1.1 The ratio $e = 0.5$ sits in the band $0 < e < 1$ . That band is the ellipse. Every point must stay closer to the focus than to the line (ratio below 1), so the curve wraps around and closes up.

Q1.2 " $P F = P M$ " means $\frac{P F}{P M} = 1$ , so $e = 1$ . That is the parabola — the exact borderline where the curve just barely fails to close.

Q1.3 "Farther from the focus than from the line" means $P F > P M$ is possible, i.e. $\frac{P F}{P M} > 1$ somewhere, so $e > 1$ . This is the hyperbola, which opens into two branches.

Level 2 — Application

Goal: plug numbers into the definition and produce the equation.

Q2.1 Find the equation of the conic with focus $F = (2, 0)$ , directrix $x = - 2$ , and $e = 1$ .

Q2.2 Find the equation of the conic with focus $F = (0, 0)$ , directrix $x = 6$ , and $e = \frac{1}{2}$ . Leave it as a tidy polynomial equation (no fractions).

Q2.3 Find the equation of the conic with focus $F = (0, 0)$ , directrix $x = 1$ , and $e = 3$ .

Recall Solution — L2

Q2.1 ( $e = 1$ , parabola) Step 1 — WHAT: write both distances. $P F = (x - 2)^{2} + y^{2}$ (distance formula to $(2, 0)$ ); $P M = ∣ x + 2∣$ (perpendicular gap to the vertical line $x = - 2$ , with $M = (- 2, y)$ ). Step 2 — apply the rule with $e = 1$ : $(x - 2)^{2} + y^{2} = ∣ x + 2∣$ . Step 3 — square. Why allowed? Both sides are genuine distances, hence $\geq 0$ ; squaring $A = B$ into $A^{2} = B^{2}$ is reversible exactly when $A, B \geq 0$ . So: $(x - 2)^{2} + y^{2} = (x + 2)^{2}$ . Step 4 — expand: $x^{2} - 4 x + 4 + y^{2} = x^{2} + 4 x + 4$ . Step 5 — cancel $x^{2}$ and $4$ : $y^{2} = 8 x$ . ✅ A parabola with $4 a = 8$ , so $a = 2$ .

Q2.2 ( $e = \frac{1}{2}$ , ellipse) Step 1: $P F = x^{2} + y^{2}$ ; $P M = ∣ x - 6∣$ (directrix is now to the right of the focus, $M = (6, y)$ ). Step 2: $x^{2} + y^{2} = \frac{1}{2} ∣ x - 6∣$ . Step 3 — square. Why allowed? Left side is $\geq 0$ and right side is $\frac{1}{2}$ times an absolute value $\geq 0$ ; both non-negative, so squaring is safe: $x^{2} + y^{2} = \frac{1}{4} (x - 6)^{2}$ . Step 4 — multiply by 4: $4 x^{2} + 4 y^{2} = x^{2} - 12 x + 36$ . Step 5 — collect: $3 x^{2} + 4 y^{2} + 12 x - 36 = 0$ . ✅ Both squared terms positive ⇒ closed ellipse.

Q2.3 ( $e = 3$ , hyperbola) Step 1: $P F = x^{2} + y^{2}$ ; $P M = ∣ x - 1∣$ ( $M = (1, y)$ ). So $x^{2} + y^{2} = 3∣ x - 1∣$ . Step 2 — square. Why allowed? Again both sides are non-negative distances, so squaring introduces no phantom solutions: $x^{2} + y^{2} = 9 (x - 1)^{2} = 9 x^{2} - 18 x + 9$ . Step 3 — collect on one side: $0 = 8 x^{2} - 18 x + 9 - y^{2}$ , i.e. $8 x^{2} - y^{2} - 18 x + 9 = 0$ . ✅ The $+ 8 x^{2}$ against $- y^{2}$ (opposite signs) ⇒ hyperbola, two branches.

Level 3 — Analysis

Goal: reverse the process — extract $e$ , foci, directrices, or classify from a raw equation.

Q3.1 A conic is defined by focus $F = (0, 0)$ , directrix $x = - 4$ , eccentricity $e$ . Its point $P = (x, y)$ satisfies $x^{2} + y^{2} = e ∣ x + 4∣$ . If the point $(0, 3)$ lies on the curve, find $e$ and classify the conic.

Q3.2 Given the standard parabola $y^{2} = 4 a x$ with $a = 5$ , state its focus, its directrix, and its eccentricity. (Uses Parabola - Standard equation y^2=4ax.)

Q3.3 For an ellipse the geometric formula $e = \frac{c}{a}$ holds, where $a$ is the semi-major axis and $c$ the centre-to-focus distance (see Ellipse - Standard form and c^2 = a^2 - b^2). If $a = 5$ and $c = 3$ , find $e$ and confirm it lies in the ellipse band.

Recall Solution — L3

Q3.1 Step 1 — plug the point $(0, 3)$ into the given relation: $0^{2} + 3^{2} = e ∣0 + 4∣$ . Step 2 — evaluate: $3 = e \cdot 4$ . Step 3 — solve: $e = \frac{3}{4} = 0.75$ . Step 4 — classify: $0 < 0.75 < 1$ ⇒ ellipse.

Q3.2 For $y^{2} = 4 a x$ the focus is at $(a, 0)$ and the directrix is $x = - a$ (this is the focus–directrix construction with $e = 1$ ). With $a = 5$ :

Focus $= (5, 0)$ .
Directrix $x = - 5$ .
A parabola is always $e = 1$ .

Q3.3 Step 1: $e = \frac{c}{a} = \frac{3}{5} = 0.6$ . Step 2: $0 < 0.6 < 1$ ✅ — safely inside the ellipse band.

Level 4 — Synthesis

Goal: combine the definition with a second condition — a point, a chord, or the latus rectum.

Q4.1 A conic has focus $F = (0, 0)$ and directrix $x = - 3$ . Find $e$ so that the point $(1, 0)$ (on the axis) is on the curve. Then classify.

Q4.2 For the parabola $y^{2} = 4 a x$ with $a = 2$ , the latus rectum is the chord through the focus perpendicular to the axis. Its full length is $4 a$ (see Latus rectum and semi-latus rectum). Find (a) the length of the latus rectum, and (b) the two endpoints of that chord.

Q4.3 A conic has focus $F = (0, 0)$ , directrix $x = - 2$ , and $e = 1$ . Derive its equation from scratch, then read off its vertex (the point on the axis of symmetry).

Recall Solution — L4

Q4.1 Step 1: $P F = 1^{2} + 0^{2} = 1$ ; $P M = ∣1 + 3∣ = 4$ (the point $M = (- 3, 0)$ is the foot of the perpendicular from $(1, 0)$ to $x = - 3$ ). Step 2: the rule $P F = e P M$ gives $1 = e \cdot 4$ . Step 3: $e = \frac{1}{4} = 0.25$ . Step 4: $0 < 0.25 < 1$ ⇒ ellipse.

Q4.2 ( $y^{2} = 4 a x$ , $a = 2$ , so $y^{2} = 8 x$ , focus $(2, 0)$ ) (a) Length of latus rectum $= 4 a = 4 (2) = 8$ . (b) The chord passes through the focus $x = 2$ : substitute into $y^{2} = 8 x$ : $y^{2} = 16$ , so $y = \pm 4$ . Endpoints $(2, 4)$ and $(2, - 4)$ . The distance between them is $4 - (- 4) = 8$ ✅ (matches $4 a$ ).

Q4.3 Step 1 — distances: $P F = x^{2} + y^{2}$ ; $P M = ∣ x + 2∣$ (foot $M = (- 2, y)$ ). Step 2 — rule with $e = 1$ : $x^{2} + y^{2} = ∣ x + 2∣$ . Step 3 — square (both sides non-negative distances, so reversible): $x^{2} + y^{2} = (x + 2)^{2} = x^{2} + 4 x + 4$ . Step 4 — cancel $x^{2}$ : $y^{2} = 4 x + 4 = 4 (x + 1)$ . Step 5 — vertex: the parabola $y^{2} = 4 (x + 1)$ has its turning point where $y = 0$ , i.e. $x = - 1$ . Vertex $= (- 1, 0)$ . (Sanity: it sits halfway between focus $(0, 0)$ and directrix $x = - 2$ ✅.)

Level 5 — Mastery

Goal: full derivations, degenerate limits, and case-splitting across quadrants/signs.

Q5.1 Starting from focus $F = (0, 0)$ and directrix $x = - d$ ( $d > 0$ ), derive the general conic $x^{2} (1 - e^{2}) - 2 e^{2} d x + y^{2} = e^{2} d^{2}$ , then show explicitly which term vanishes when $e = 1$ and why that forces a parabola.

Q5.2 (Degenerate limit) Investigate what happens to the focus–directrix rule as $e \to 0$ . Explain, using the algebra, why the curve collapses to a circle centred at the focus, connecting to Circle as limiting case e=0.

Q5.3 (Case-split on directrix side) A conic has focus $F = (0, 0)$ and $e = 2$ . Write its equation once with directrix $x = 1$ (to the right of the focus) and once with directrix $x = - 1$ (to the left). Show the two equations and explain the sign difference in the linear term.

Q5.4 (Hyperbola classification with signs) Classify $8 x^{2} - y^{2} - 18 x + 9 = 0$ (from Q2.3) by the sign test, and explain in one line why one variable can grow without bound.

Recall Solution — L5

Q5.1 Step 1 — distances: $P F = x^{2} + y^{2}$ , $P M = ∣ x + d ∣$ (foot $M = (- d, y)$ ). Step 2 — rule: $x^{2} + y^{2} = e ∣ x + d ∣$ . Step 3 — square (both sides non-negative distances): $x^{2} + y^{2} = e^{2} (x + d)^{2}$ . Step 4 — expand the right: $e^{2} (x + d)^{2} = e^{2} x^{2} + 2 e^{2} d x + e^{2} d^{2}$ . Step 5 — move everything left: $x^{2} - e^{2} x^{2} + y^{2} - 2 e^{2} d x = e^{2} d^{2}$ , i.e. $x^{2} (1 - e^{2}) - 2 e^{2} d x + y^{2} = e^{2} d^{2}$ . ✅ Step 6 — set $e = 1$ : the coefficient $(1 - e^{2}) = 0$ , so the $x^{2}$ term vanishes, leaving $- 2 d x + y^{2} = d^{2}$ , i.e. $y^{2} = 2 d x + d^{2}$ — only a first power of $x$ survives, which is exactly the signature of a parabola.

Q5.2 In the rule $P F = e P M$ , as $e \to 0$ the right side $\to 0$ , forcing $P F \to 0$ : every point collapses onto the focus. To see a proper circle instead, hold the size fixed while pushing the directrix away. Start from the expanded right-hand side: $e^{2} (x + d)^{2} = e^{2} d^{2} + 2 e^{2} d x + e^{2} x^{2} .$ Now let $d \to \infty$ while keeping $e^{2} d^{2} \to R^{2}$ (a finite constant). Then:

the constant term $e^{2} d^{2} \to R^{2}$ ;
the linear coefficient $2 e^{2} d = \frac{2 ( e ^{2} d ^{2} )}{d} \to \frac{2 R ^{2}}{d} \to 0$ (numerator fixed, denominator $\to \infty$ );
the quadratic coefficient $e^{2} = \frac{e ^{2} d ^{2}}{d ^{2}} \to \frac{R ^{2}}{d ^{2}} \to 0$ (numerator fixed, denominator $\to \infty$ ).

So the whole right-hand side reduces to just $R^{2}$ , and $x^{2} + y^{2} = e^{2} (x + d)^{2}$ becomes $x^{2} + y^{2} = R^{2}$ — a circle of radius $R$ centred at the focus. That is why $e = 0$ is the round, no-directrix limit.

Q5.3 Directrix $x = 1$ : $P M = ∣ x - 1∣$ . Then $x^{2} + y^{2} = 2∣ x - 1∣$ ⇒ (squaring both non-negative sides) $x^{2} + y^{2} = 4 (x - 1)^{2} = 4 x^{2} - 8 x + 4$ ⇒ collect: $3 x^{2} - y^{2} - 8 x + 4 = 0$ . Directrix $x = - 1$ : $P M = ∣ x + 1∣$ . Then $x^{2} + y^{2} = 2∣ x + 1∣$ ⇒ $x^{2} + y^{2} = 4 (x + 1)^{2} = 4 x^{2} + 8 x + 4$ ⇒ $3 x^{2} - y^{2} + 8 x + 4 = 0$ . The only difference is the sign of the linear $x$ -term ( $- 8 x$ vs $+ 8 x$ ): shifting the directrix from the right to the left of the focus mirrors the whole curve across the $y$ -axis, which flips the sign of the odd-power term while leaving the even powers ( $x^{2}, y^{2}$ , constant) untouched. Both are hyperbolas ( $3 x^{2}$ vs $- y^{2}$ ).

Q5.4 Group the squared terms: $8 x^{2}$ (positive) and $- y^{2}$ (negative) — opposite signs ⇒ hyperbola. Because $y^{2}$ is subtracted, we can hold the equation balanced while $y^{2}$ grows as fast as $8 x^{2}$ does; so as $x \to \infty$ , $y$ grows without bound too — the branch runs off to infinity instead of closing.

Recall Final self-check — answer without peeking, then reveal

Focus $(0, 0)$ , directrix $x = - 3$ , point $(1, 0)$ on curve — find $e$ and classify. ::: $P F = 1$ , $P M = 4$ , so $e = \frac{1}{4} = 0.25$ ; since $0 < 0.25 < 1$ it is an ellipse. Length of the latus rectum of $y^{2} = 8 x$ . ::: Here $4 a = 8$ , so the length is $8$ . Which term vanishes at $e = 1$ , and why. ::: The $(1 - e^{2}) x^{2}$ term, because $1 - e^{2} = 0$ when $e = 1$ , leaving only a first power of $x$ (a parabola). The band of $e$ that names a hyperbola. ::: $e > 1$ . Perpendicular distance from $(x, y)$ to the vertical line $x = k$ . ::: $∣ x - k ∣$ .

Connections

Parent: 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

Exercises — Definition via focus, directrix, eccentricity e