Conic Sections

⚡ Key Takeaways

Conic sections are the curves obtained by intersecting a plane with a double cone: the circle, ellipse, parabola, and hyperbola. Their key unifying property is that each is the locus of points whose distance to a fixed point (focus) and a fixed line (directrix) maintains a constant ratio called the eccentricity $e$ . They appear everywhere from planetary orbits to satellite dishes to projectile motion. Studied systematically by Apollonius of Perga (~200 BCE), they became the language of celestial mechanics with Kepler and Newton.

💡 The Core Insight

Every conic is governed by eccentricity $e \ge 0$ , the constant ratio of a point's distance from a fixed focus to its distance from a fixed directrix. When $e = 0$ you get a circle, $0 < e < 1$ an ellipse, $e = 1$ a parabola, and $e > 1$ a hyperbola. One definition, one parameter, four curves. The geometry of slicing a cone and the algebra of $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ are two faces of this single idea.

DefinitionConic Section — Geometric (Slicing) Definition

A conic section is the intersection of a plane with a right circular double cone. Let the cone have half-angle $\alpha$ (angle between axis and a generating line), and let the cutting plane make angle $\beta$ with the cone's axis. The type of conic depends on the comparison of $\beta$ and $\alpha$ :

1

\quad \text{conic type} = \begin{cases} \text{ellipse (circle)} & \beta > \alpha \\ \text{parabola} & \beta = \alpha \\ \text{hyperbola} & \beta < \alpha \end{cases}

Derivation hint: A circle results when the plane is perpendicular to the axis ( $\beta = 90°$ ); tilting the plane until it becomes parallel to a generator ( $\beta = \alpha$ ) produces the parabola; tilting further so it meets both nappes yields the hyperbola.

DefinitionConic Section — Focus-Directrix Definition

A conic is the locus of points $P$ such that the ratio of the distance from $P$ to a fixed point $F$ (the focus) to the distance from $P$ to a fixed line $\ell$ (the directrix) equals a constant $e \ge 0$ (the eccentricity):

2

\quad \frac{|PF|}{\text{dist}(P,\ell)} = e

Here $P$ is a variable point on the curve, $F$ is the focus, $\ell$ is the directrix, and $e$ is the eccentricity. The classification is: $e=0$ circle, $0<e<1$ ellipse, $e=1$ parabola, $e>1$ hyperbola. Derivation hint: Set $F=(0,0)$ , let $\ell$ be the vertical line $x=-d$ , write $|PF|=\sqrt{x^2+y^2}$ and $\text{dist}(P,\ell)=|x+d|$ , then square the ratio equation to obtain a second-degree polynomial.

TheoremGeneral Second-Degree Equation

Every conic section is the solution set of a second-degree equation in two variables:

3

\quad Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

with $A,B,C$ not all zero. The conic type is determined by the discriminant $B^2 - 4AC$ :

4

\quad \text{type} = \begin{cases} \text{ellipse} & B^2 - 4AC < 0 \\ \text{parabola} & B^2 - 4AC = 0 \\ \text{hyperbola} & B^2 - 4AC > 0 \end{cases}

Derivation hint: The discriminant is invariant under rotation of axes; rotate to eliminate the $Bxy$ term, reducing to standard form whose sign structure reveals the type.

DefinitionStandard Forms (Axes-Aligned, Centered/Vertexed)

With center or vertex chosen at convenient points and axes aligned to coordinate axes:

5

\quad \text{Circle:}\quad x^2 + y^2 = r^2

Derivation hint: Special case of the ellipse with $a=b=r$ ; all points at fixed distance $r$ from the center.

6

\quad \text{Ellipse:}\quad \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b > 0

Derivation hint: Sum of distances to two foci is constant ( $=2a$ ); algebraically simplify $|PF_1|+|PF_2|=2a$ .

7

\quad \text{Parabola:}\quad y^2 = 4px

Derivation hint: From the focus-directrix definition with $e=1$ , focus $(p,0)$ , directrix $x=-p$ .

8

\quad \text{Hyperbola:}\quad \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

Derivation hint: Absolute difference of distances to two foci is constant ( $=2a$ ); simplify $\big||PF_1|-|PF_2|\big|=2a$ .

DefinitionKey Parameters of Ellipse and Hyperbola

For the ellipse and hyperbola, the focal distance $c$ relates the semi-axes:

9

\quad \text{Ellipse:}\ c^2 = a^2 - b^2, \qquad \text{Hyperbola:}\ c^2 = a^2 + b^2

Derivation hint: For the ellipse, evaluate the sum-of-distances at the end of the minor axis where each focal distance equals $a$ ; Pythagoras gives $b^2 + c^2 = a^2$ . The eccentricity in terms of axes is:

10

\quad e = \frac{c}{a}

Derivation hint: Substitute the focus location and directrix back into the ratio definition; for the ellipse $e=c/a<1$ , for the hyperbola $e=c/a>1$ .

ProofDeriving the Ellipse from the Two-Focus Definition

We show that "sum of distances to two foci is constant" yields equation [6]. 1. Place foci at $F_1 = (-c, 0)$ and $F_2 = (c, 0)$ . Let the constant sum be $2a$ with $a > c > 0$ . 2. For a point $P=(x,y)$ on the curve:

11

\quad \sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2 + y^2} = 2a

Derivation hint: Direct statement of the defining locus condition. 3. Isolate one radical and square:

12

\quad \sqrt{(x+c)^2+y^2} = 2a - \sqrt{(x-c)^2+y^2}

Squaring both sides:

(x+c)^2 + y^2 = 4a^2 - 4a\sqrt{(x-c)^2+y^2} + (x-c)^2 + y^2

Derivation hint: Expand both squared binomials; the $y^2$ and $x^2,c^2$ terms partly cancel. 4. Expanding and cancelling $x^2, y^2, c^2$ leaves $4cx = 4a^2 - 4a\sqrt{(x-c)^2+y^2}$ , so:

13

\quad a\sqrt{(x-c)^2+y^2} = a^2 - cx

Derivation hint: Divide through by 4 and rearrange to isolate the remaining radical. 5. Square again:

a^2\big((x-c)^2 + y^2\big) = a^4 - 2a^2 cx + c^2 x^2

Expand and group:

14

\quad (a^2 - c^2)x^2 + a^2 y^2 = a^2(a^2 - c^2)

Derivation hint: Collect $x^2$ terms; the linear $x$ terms cancel cleanly. 6. Define $b^2 = a^2 - c^2$ (positive since $a>c$ ). Divide by $a^2 b^2$ :

15

\quad \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad \blacksquare

Derivation hint: Substitute $a^2-c^2=b^2$ into [14] and normalize the right side to 1.

∎

ProofParabola from Focus-Directrix with

e=1

1. Focus $F=(p,0)$ , directrix $\ell: x = -p$ . Locus condition $|PF| = \text{dist}(P,\ell)$ :

16

\quad \sqrt{(x-p)^2 + y^2} = |x + p|

Derivation hint: Equal distances since $e=1$ ; right side is horizontal distance to the line. 2. Square both sides:

(x-p)^2 + y^2 = (x+p)^2

Derivation hint: Both sides nonnegative, so squaring preserves equality. 3. Expand: $x^2 - 2px + p^2 + y^2 = x^2 + 2px + p^2$ . Cancel $x^2, p^2$ :

17

\quad y^2 = 4px \qquad \blacksquare

Derivation hint: The $-2px$ and $+2px$ combine to give $y^2 = 4px$ .

∎

Figure

The four conic sections as slices of a double cone, plus the ellipse in the plane

mathematician · Apollonius of Perga | c. 262 – c. 190 BCE

The Greek geometer Apollonius wrote Conics, an eight-book treatise that systematized the study of these curves. He coined the names ellipse ("falling short"), parabola ("alongside"), and hyperbola ("exceeding") — terms borrowed from the Pythagorean theory of application of areas, describing whether a constructed rectangle fell short of, matched, or exceeded a reference area. Crucially, Apollonius showed that all three curves arise from a single cone by varying the cutting plane, rather than requiring three differently-shaped cones as earlier geometers (Menaechmus) had supposed. His work remained the definitive reference for nearly 2000 years and directly enabled Kepler and Newton.

🧠 Intuition

Imagine shining a flashlight (a cone of light) at a wall.

Point it straight on: a circle.
Tilt it slightly: the circle stretches into an ellipse — still a closed loop because the cone "closes" around the beam.
Tilt until one edge of the light beam is parallel to the wall: the loop never closes and you get an open parabola.
Tilt further so the beam would, if extended backward, hit the wall on the other side too — you reveal both branches of a hyperbola.

Eccentricity measures "how stretched": $e=0$ is perfectly round; as $e\to 1$ the ellipse becomes more and more elongated until it bursts open into a parabola; beyond that ( $e>1$ ) it splits into two branches racing toward straight-line asymptotes.

ExampleIdentifying and Analyzing an Ellipse

Given: The equation $9x^2 + 25y^2 = 225$ . Find the type, semi-axes, foci, and eccentricity.

Steps: 1. Divide by 225 to reach standard form:

\frac{x^2}{25} + \frac{y^2}{9} = 1

2. Read off $a^2 = 25 \Rightarrow a = 5$ (major, along $x$ ) and $b^2 = 9 \Rightarrow b = 3$ . 3. Discriminant check ([4]): $A=9, B=0, C=25 \Rightarrow B^2-4AC = -900 < 0$ → ellipse. ✓ 4. Focal distance from [9]: $c^2 = a^2 - b^2 = 25 - 9 = 16 \Rightarrow c = 4$ . Foci at $(\pm 4, 0)$ . 5. Eccentricity from [10]: $e = c/a = 4/5 = 0.8$ .

Answer:

\boxed{a=5,\quad b=3,\quad \text{foci }(\pm 4,\,0),\quad e = 0.8}

Since $0<e<1$ this confirms an ellipse, fairly elongated.

ExampleFinding a Parabola's Focus and Directrix

Given: A satellite dish has cross-section $y^2 = 12x$ (units in cm). Where should the receiver be placed?

Steps: 1. Compare with standard form [7]: $y^2 = 4px$ . So $4p = 12 \Rightarrow p = 3$ . 2. The focus of $y^2 = 4px$ sits at $(p, 0)$ , so focus $= (3, 0)$ cm. 3. The directrix is $x = -p = -3$ cm. 4. Physical meaning: incoming parallel rays (signals) reflect off the parabola and all converge at the focus — that's where the receiver goes.

Answer:

\boxed{\text{Receiver at } (3,\,0)\ \text{cm; directrix } x=-3\ \text{cm}}

The receiver sits 3 cm from the vertex along the axis.

⚠️ Common mistake

A circle is not "the same as" an ellipse with $a=b$ in the focus picture. While $x^2+y^2=r^2$ is indeed the ellipse limit as $b \to a$ , the two foci merge into the single center and the directrix recedes to infinity. So the focus-directrix ratio [2] is a limiting case ( $e=0$ ), not literally evaluable as $0/0$ at finite scale. Treat the circle's $e=0$ as a definition reached by limit, not by plugging into the ratio.

The sign of the $xy$ term matters, not just $A$ and $C$ . A common error is classifying $x^2 + y^2 + xy = 1$ as an ellipse just because $A,C>0$ . You must use the full discriminant: $B^2 - 4AC = 1 - 4 = -3 < 0$ → ellipse here (correct), but $x^2 + 4xy + y^2 = 1$ gives $16 - 4 = 12 > 0$ → hyperbola, despite "looking" elliptical.

Counterexample

Not every second-degree equation is a "nice" conic. Consider $x^2 - y^2 = 0$ . Discriminant $B^2-4AC = 0 - 4(1)(-1) = 4 > 0$ suggests a hyperbola, but factoring gives $(x-y)(x+y)=0$ — a pair of intersecting lines, a degenerate hyperbola. Similarly $x^2 + y^2 = 0$ is just the single point $(0,0)$ (degenerate ellipse), and $x^2 = 0$ is the doubled line $x=0$ (degenerate parabola). The discriminant tells you the family but not whether the conic has degenerated.

connection

Conic sections sit at a crossroads of mathematics. Algebraically they are the level sets of Quadratic Forms in two variables, and rotating axes to eliminate the $xy$ term is an application of Eigenvalues and Eigenvectors (the principal axes are eigenvectors of the coefficient matrix). In Polar Coordinates all four conics share one beautiful equation $r = \frac{\ell}{1 + e\cos\theta}$ with the focus at the origin — the form Kepler needed. They generalize to Quadric Surfaces in three dimensions (ellipsoids, paraboloids, hyperboloids) and are unified projectively in Projective Geometry, where ellipse, parabola, and hyperbola become the same curve viewed differently. Their reflective properties tie directly to Geometric Optics.

🔬 Where this appears

⏳ History

- ~350 BCE — Menaechmus introduces conics to solve the problem of doubling the cube, using sections of three different cones.

- ~300 BCE — Euclid writes a (lost) treatise on conics, consolidating earlier Greek work.

- ~200 BCE — Apollonius of Perga composes Conics; names ellipse, parabola, hyperbola; shows all arise from a single cone.

- ~350 CE — Pappus of Alexandria records the focus-directrix property, foreshadowing the unified eccentricity definition.

- 1609 — Johannes Kepler publishes Astronomia Nova, establishing that Mars (and planets) move in ellipses with the Sun at a focus.

- 1637 — René Descartes & Pierre de Fermat independently develop coordinate geometry, recasting conics as second-degree equations

Ax^2+Bxy+\dots=0

.

- 1655 — John Wallis publishes Tractatus de Sectionibus Conicis, treating conics purely algebraically for the first time.

- 1687 — Isaac Newton derives Kepler's elliptical orbits from the inverse-square law of gravitation in Principia, explaining why orbits are conics.

- 1822 — Germinal Pierre Dandelin proves the Dandelin spheres theorem, elegantly linking the slicing definition to the focus-directrix definition.

open-problem

The focus-directrix definition [2] cleanly produces the parabola ( $e=1$ ) but treats the circle ( $e=0$ ) only as a limit — there is no finite directrix. A subtle question is how to define conics uniformly: the projective viewpoint solves this by adjoining a "line at infinity," whereupon parabola, ellipse, and hyperbola differ only in how they meet that line (tangent, miss, cross). Yet teaching this rigorously to beginners while preserving the metric (distance-based) intuition remains pedagogically open. A second subtlety: classifying the degenerate cases (point, line, two lines) requires examining a $3\times 3$ determinant beyond the simple discriminant $B^2-4AC$ , and the boundary between "true" and "degenerate" conics is a recurring source of student error. Finally, generalizations to higher dimensions (Quadric Surfaces) introduce many more types, and the full classification over the complex numbers versus the reals differs in ways often glossed over at the introductory level.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also