Conic sections · The Lumenary

MotivationH

Curves with a secret point

A solar cooker is a shiny dish that gathers sunlight onto a pot. Families in sunny regions from India to Bolivia use them to cook without firewood. The dish is not just any bowl: its cross-section is a parabola, and every ray of sunlight that enters parallel to its axis is reflected through a single point, the focus. Satellite dishes and radio telescopes work the same way, in reverse for car headlights.

In some hospitals, kidney stones are shattered without surgery by a lithotripter: a shock wave is produced at one focus of a reflector shaped like part of an ellipsoid, and every reflected wave arrives together at the other focus — where the stone has been placed. And when Johannes Kepler studied the orbit of Mars, he found it was an ellipse with the Sun at one focus. In 2017 astronomers tracked the object ʻOumuamua passing the Sun on a hyperbola: it came from interstellar space and is leaving for good.

Parabolas, ellipses and hyperbolas are the conic sections: the curves you get by slicing a cone. In this lesson we meet three descriptions of them — slices of a cone, sets of points defined by distances, and second-degree equations — and prove that they agree.

IntuitionH

Slicing a cone

Take a double cone: two ice-cream cones joined at their tips, extending forever. Slice it with a flat plane.

A plane perpendicular to the axis gives a circle.
Tilt the plane a little: the slice is a closed oval, an ellipse.
Tilt it until it is parallel to a line on the cone (a generator): the curve no longer closes up. It is a parabola.
Tilt it further, so that it cuts both halves of the cone: you get the two branches of a hyperbola.
A plane through the tip gives a point, a single line or a pair of crossing lines: the degenerate conics.

A side view of the double cone and four cutting planes (seen edge-on). The horizontal plane gives a circle, the tilted one an ellipse, the one parallel to a generator a parabola, and the steep one, which cuts both halves, a hyperbola.

Slices are hard to compute with, so the ancient Greek geometers also described the curves by distances. You can draw an ellipse with two pins and a loop of string: put the pins at two points, keep the string taut with a pencil, and move round. The total length from the pencil to the two pins never changes. Gardeners lay out oval flower beds this way.

DefinitionsH

Definitions

Definition— ellipse

Given two points $F_1$ , $F_2$ (the foci) and a length $2a$ greater than the distance $F_1F_2$ , the ellipse is the set of points $P$ with

PF_1 + PF_2 = 2a.

Definition— parabola

Given a point $F$ (the focus) and a line $\ell$ (the directrix) not through $F$ , the parabola is the set of points equidistant from $F$ and from $\ell$ .

Definition— hyperbola

Given two foci $F_1$ , $F_2$ and a length $2a$ with $0 \lt 2a \lt F_1F_2$ , the hyperbola is the set of points $P$ with

\left|PF_1 - PF_2\right| = 2a.

Definition— eccentricity

For a point $F$ , a line $\ell$ not through $F$ , and a number $e > 0$ , the set of points $P$ with $PF = e \cdot (\text{distance from } P \text{ to } \ell)$ is an ellipse if $e \lt 1$ , a parabola if $e = 1$ and a hyperbola if $e > 1$ (we prove this below). The number $e$ is the eccentricity. A circle is the limiting case $e = 0$ ; it has no directrix.

Placing the foci symmetrically on the $x$ -axis gives the standard equations. In each row, $a$ , $b$ , $c$ are positive (for the parabola, $p \ne 0$ ):

Curve	Foci	Equation	Relation	Eccentricity
Ellipse	$(\pm c, 0)$	$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$	$b^2 = a^2 - c^2$	$e = c/a \lt 1$
Hyperbola	$(\pm c, 0)$	$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$	$b^2 = c^2 - a^2$	$e = c/a > 1$
Parabola	$(0, p)$ ; directrix $y = -p$	$x^2 = 4py$	—	$e = 1$

For the ellipse, $a$ is the semi-major axis and $b$ the semi-minor axis. The hyperbola has asymptotes $y = \pm\tfrac{b}{a}x$ : its branches get closer and closer to these lines without meeting them. The ellipse with $a = b$ (so $c = 0$ ) is the circle $x^2 + y^2 = a^2$ .

Worked examplesH

Worked examples

Example— The equation of a parabola from its focus and directrix

Take the focus $F = (0, 2)$ and the directrix $y = -2$ . A point $P = (x, y)$ is at distance $\sqrt{x^2 + (y - 2)^2}$ from $F$ and $|y + 2|$ from the line. Setting them equal and squaring: $x^2 + y^2 - 4y + 4 = y^2 + 4y + 4$ , so $x^2 = 8y$ , or $y = \tfrac18 x^2$ . In general, focus $(0, p)$ and directrix $y = -p$ give $x^2 = 4py$ .

Example— Finding the foci of an ellipse

$9x^2 + 25y^2 = 225$ . Divide by $225$ : $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ , so $a = 5$ , $b = 3$ and $c = \sqrt{25 - 9} = 4$ . The foci are $(\pm 4, 0)$ , the eccentricity is $\tfrac45$ , and every point of the ellipse has distances to the foci adding up to $2a = 10$ . Check at the top point $(0, 3)$ : both distances are $\sqrt{16 + 9} = 5$ .

Example— Completing the square

What is $4x^2 + 9y^2 - 16x + 54y + 61 = 0$ ? Group and complete the squares (see quadratics):

4(x^2 - 4x) + 9(y^2 + 6y) + 61 = 4(x - 2)^2 - 16 + 9(y + 3)^2 - 81 + 61,

so the equation is $4(x - 2)^2 + 9(y + 3)^2 = 36$ , that is,

\frac{(x - 2)^2}{9} + \frac{(y + 3)^2}{4} = 1:

an ellipse with centre $(2, -3)$ , $a = 3$ , $b = 2$ and foci $(2 \pm \sqrt5, -3)$ . If the constant had been $97$ instead of $61$ , we would have reached $4(x - 2)^2 + 9(y + 3)^2 = 0$ , a single point; with a larger constant, no points at all.

Example— A hyperbola in disguise

The graph of $y = 1/x$ , that is $xy = 1$ , is a hyperbola whose asymptotes are the coordinate axes. Rotating it by $45^\circ$ (substitute $x = (X - Y)/\sqrt2$ , $y = (X + Y)/\sqrt2$ ) turns $xy = 1$ into $X^2 - Y^2 = 2$ , which is in standard form with $a = b = \sqrt2$ and $c = 2$ . The $xy$ term was hiding the standard form.

Example— Non-examples

$x^2 - y^2 = 0$ factorises as $(x - y)(x + y) = 0$ : a pair of crossing lines, not a hyperbola.
$x^2 + y^2 = 0$ is a single point, and $x^2 + y^2 + 1 = 0$ has no real points at all.
A parabola is not half of a hyperbola and has no asymptotes: on $y = x^2/4$ the slope $x/2$ grows without bound, so no straight line $y = mx + k$ stays close to it (the gap $x^2/4 - mx - k$ tends to infinity).

ProofH

Why the descriptions agree

Theorem— Theorem (the parabola)

A point $(x, y)$ is equidistant from the focus $(0, p)$ and the directrix $y = -p$ if and only if $x^2 = 4py$ .

Proof

Both distances, $\sqrt{x^2 + (y - p)^2}$ and $|y + p|$ , are non-negative, so they are equal exactly when their squares are: $x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2$ , which simplifies to $x^2 = 4py$ . $\blacksquare$

∎

Theorem— Theorem (the ellipse)

Let $0 \le c \lt a$ and $b^2 = a^2 - c^2$ . A point $P = (x, y)$ satisfies $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ if and only if $PF_1 + PF_2 = 2a$ , where $F_1 = (c, 0)$ and $F_2 = (-c, 0)$ . Moreover, on the ellipse

PF_1 = a - \frac{c}{a}x \qquad\text{and}\qquad PF_2 = a + \frac{c}{a}x.

Proof

Write $d_1 = PF_1 = \sqrt{(x - c)^2 + y^2}$ and $d_2 = PF_2 = \sqrt{(x + c)^2 + y^2}$ .

Equation $\Rightarrow$ distances. On the ellipse, $y^2 = b^2\left(1 - \frac{x^2}{a^2}\right) = (a^2 - c^2)\left(1 - \frac{x^2}{a^2}\right)$ . Substituting,

d_1^2 = x^2 - 2cx + c^2 + a^2 - c^2 - x^2 + \frac{c^2x^2}{a^2} = a^2 - 2cx + \frac{c^2 x^2}{a^2} = \left(a - \frac{c}{a}x\right)^2.

Since $|x| \le a$ on the ellipse and $c \lt a$ , the number $a - \frac{c}{a}x$ is positive, so $d_1 = a - \frac{c}{a}x$ . Replacing $x$ by $-x$ gives $d_2 = a + \frac{c}{a}x$ , and $d_1 + d_2 = 2a$ .

Distances $\Rightarrow$ equation. Suppose $d_1 + d_2 = 2a$ . Subtracting the squares, $d_2^2 - d_1^2 = (x + c)^2 - (x - c)^2 = 4cx$ . Since $d_2^2 - d_1^2 = (d_2 - d_1)(d_2 + d_1) = 2a(d_2 - d_1)$ , we get $d_2 - d_1 = \frac{2cx}{a}$ . Together with $d_1 + d_2 = 2a$ this gives $d_1 = a - \frac{c}{a}x$ . Squaring, $(x - c)^2 + y^2 = a^2 - 2cx + \frac{c^2x^2}{a^2}$ , which rearranges to $x^2\left(1 - \frac{c^2}{a^2}\right) + y^2 = a^2 - c^2$ . Dividing by $a^2 - c^2 = b^2$ gives $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ . $\blacksquare$

∎

The same algebra, with $b^2 = c^2 - a^2$ , shows that $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is the hyperbola $|d_1 - d_2| = 2a$ : on its right branch $d_1 = \frac{c}{a}x - a$ and $d_2 = \frac{c}{a}x + a$ .

Theorem— Corollary (focus and directrix of an ellipse)

If $c > 0$ , put $e = c/a$ . Every point of the ellipse satisfies $PF_1 = e \cdot (\text{distance from } P \text{ to the line } x = a/e)$ .

Proof

$PF_1 = a - \frac{c}{a}x = e\left(\frac{a}{e} - x\right)$ , and $\frac{a}{e} - x$ is the distance from $P$ to the vertical line $x = a/e$ , because $x \le a \lt a/e$ . $\blacksquare$

∎

Theorem— Theorem (eccentricity decides the type)

Let $F$ be the origin, let $\ell$ be the line $x = -k$ with $k > 0$ , and let $e > 0$ . The set of points with $PF = e \cdot (\text{distance to } \ell)$ is an ellipse if $e \lt 1$ , a parabola if $e = 1$ , and a hyperbola if $e > 1$ .

Proof

Squaring $\sqrt{x^2 + y^2} = e\,|x + k|$ gives

(1 - e^2)\,x^2 - 2e^2k\,x + y^2 = e^2k^2.

If $e = 1$ , the $x^2$ term vanishes and $y^2 = 2kx + k^2$ : a parabola (opening to the right). If $e \ne 1$ , complete the square in $x$ , with $h = \frac{e^2k}{1 - e^2}$ : the equation becomes $(1 - e^2)(x - h)^2 + y^2 = r$ with $r = \frac{e^2k^2}{1 - e^2}$ . If $e \lt 1$ , both squares have positive coefficients and $r > 0$ : an ellipse. If $e > 1$ , the coefficients have opposite signs and $r \lt 0$ ; multiplying by $-1$ gives the standard form of a hyperbola. $\blacksquare$

∎

Theorem— Theorem (Dandelin spheres: slices of a cone are ellipses)

If a plane cuts every generator of one half of a cone (and misses the tip), the curve of intersection is an ellipse.

Proof

Inscribe two spheres in the cone, one on each side of the plane, each touching the cone along a circle ( $C_1$ and $C_2$ ) and touching the plane at a single point ( $F_1$ and $F_2$ ). Let $P$ be a point of the curve. The generator through $P$ meets $C_1$ at $Q_1$ and $C_2$ at $Q_2$ . Two tangent segments drawn from a point to a sphere have equal length, so $PF_1 = PQ_1$ (both tangent to the first sphere) and $PF_2 = PQ_2$ . Hence

PF_1 + PF_2 = PQ_1 + PQ_2 = Q_1Q_2,

and $Q_1Q_2$ — the distance between the two circles measured along a generator — is the same for every generator. So the sum of distances to $F_1$ and $F_2$ is constant: the curve is an ellipse, and the points where the spheres touch the plane are its foci. $\blacksquare$

∎

Germinal Pierre Dandelin found this argument in 1822. Similar arguments (with one sphere, or with spheres in both halves of the cone) handle the parabola and the hyperbola.

ProofH

The reflective properties

Light reflects from a curved mirror as it would from the tangent line at the point where it hits: the reflected ray is the mirror image, in the tangent line, of the ray that would have continued straight on.

Theorem— Reflective property of the parabola

A ray travelling parallel to the axis of a parabola is reflected through the focus.

Proof

Let the ray come straight down and hit the parabola at $P$ . Let $D$ be the foot of the perpendicular from $P$ to the directrix, so the ray, continued straight on, would travel from $P$ towards $D$ . By definition $PF = PD$ . Let $t$ be the perpendicular bisector of $FD$ ; it passes through $P$ , because $P$ is equidistant from $F$ and $D$ .

We claim $t$ is the tangent at $P$ . Take any other point $Q$ of $t$ . Then $QF = QD$ (as $Q$ is on the perpendicular bisector), and $QD$ is longer than the distance from $Q$ to the directrix, because $QD$ is not perpendicular to the directrix (only $PD$ is). So $QF$ is greater than the distance from $Q$ to the directrix, and $Q$ is not on the parabola. Thus $t$ meets the parabola only at $P$ and is not parallel to the axis (it is perpendicular to $FD$ , which is not vertical unless $P$ is the vertex, where $t$ is the horizontal tangent): it is the tangent line.

Reflection in $t$ fixes $P$ and swaps $D$ and $F$ (that is what "perpendicular bisector of $FD$ " means). So the mirror image of the ray from $P$ towards $D$ is the ray from $P$ towards $F$ : the reflected ray passes through the focus. $\blacksquare$

∎

The parabola y = x²/4 with focus F and directrix. The vertical ray hits P; the tangent at P is the perpendicular bisector of FD, so the reflected ray goes to F — the mirror image of the dashed straight-on path to D.

Theorem— Reflective property of the ellipse

A ray from one focus of an ellipse is reflected through the other focus.

Proof

Let the ray from $F_1$ hit the ellipse at $P$ , and let $t$ be the tangent at $P$ . Because the ellipse is an oval curve (convex), every point $Q \ne P$ of $t$ lies outside it, so $QF_1 + QF_2 > 2a = PF_1 + PF_2$ . Thus $P$ is the point of $t$ with the smallest total distance to $F_1$ and $F_2$ .

Now use Heron's trick: reflect $F_2$ in $t$ to get $F_2'$ . For every $Q$ on $t$ , $QF_2 = QF_2'$ , so $QF_1 + QF_2 = QF_1 + QF_2'$ , which is smallest when $Q$ lies on the straight segment $F_1F_2'$ . So $P$ is on that segment, and the angles that $PF_1$ and $PF_2'$ make with $t$ are vertically opposite, hence equal. By the symmetry of the reflection, the angle between $t$ and $PF_2'$ equals the angle between $t$ and $PF_2$ . So $PF_1$ and $PF_2$ make equal angles with the tangent — which is exactly the law of reflection. $\blacksquare$

∎

Note

The hyperbola has a matching property: a ray aimed at one focus is reflected, from the outside of the near branch, towards the other focus. Large telescopes of the Cassegrain design combine a parabolic main mirror with a hyperbolic secondary mirror for this reason.

ProofH

Second-degree equations

Every conic in the plane has an equation of the form

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,

with $A$ , $B$ , $C$ not all zero. Conversely, which sets does such an equation describe?

Theorem— Classification of conics

The set of real solutions of $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ is one of: an ellipse (possibly a circle), a parabola, a hyperbola, or a degenerate case — a single point, one line, two parallel lines, two crossing lines, or the empty set. When the curve is non-degenerate, it is

an ellipse if $B^2 - 4AC \lt 0$ ,
a parabola if $B^2 - 4AC = 0$ ,
a hyperbola if $B^2 - 4AC > 0$ .

Proof sketch. (1) Rotating the axes through an angle $\theta$ with $\cot 2\theta = (A - C)/B$ (or $\theta = 45^\circ$ if $A = C$ ) removes the $xy$ term; a computation shows that $B^2 - 4AC$ does not change under rotation. (2) Without the $xy$ term the discriminant is $-4AC$ , and completing the square in $x$ and $y$ (or in just one of them, if $A$ or $C$ is zero) gives a standard form or one of the degenerate cases. The heart of the matter is the next pair of facts, which do not need any rotation. $\square$

Theorem— Lemma (definite and indefinite)

If $A > 0$ and $B^2 - 4AC \lt 0$ , then $Ax^2 + Bxy + Cy^2 > 0$ for every $(x, y) \ne (0, 0)$ .
If $B^2 - 4AC > 0$ , then $Ax^2 + Bxy + Cy^2$ takes both positive and negative values.

Proof

Complete the square: $4A\left(Ax^2 + Bxy + Cy^2\right) = (2Ax + By)^2 + (4AC - B^2)\,y^2$ . Both terms are non-negative, and they are both zero only if $y = 0$ and then $x = 0$ .
If $A \ne 0$ , the value at $(1, 0)$ is $A$ , and the value at $(-B, 2A)$ is $AB^2 - 2AB^2 + 4A^2C = A(4AC - B^2)$ , which has the opposite sign. If $A = 0$ , then $B \ne 0$ , and on the line $y = 1$ the value $Bx + C$ takes every real number. $\blacksquare$

∎

So when $B^2 - 4AC \lt 0$ the quadratic part grows in every direction, and the curve is bounded: an ellipse, a point or nothing. When $B^2 - 4AC > 0$ the quadratic part is zero along two directions — the asymptotes of a hyperbola, or the two lines of a degenerate case. The borderline case $B^2 - 4AC = 0$ gives a parabola (or, degenerately, two parallel lines, one line or nothing). For the eccentricity family above, $A = 1 - e^2$ , $B = 0$ , $C = 1$ , so $B^2 - 4AC = 4(e^2 - 1)$ : negative for $e \lt 1$ , zero for $e = 1$ , positive for $e > 1$ , just as it should be.

Formal mathematicsF

Conics in Lean

Mathlib has no special theory of conic sections, so everything here is done in coordinates, with real numbers. First the parabola as a locus — the distance to the focus (0, p) equals the distance |y + p| to the directrix:

parabola_focus_directrixLean 4Proved here from basic factsFormally verified

/-- **The parabola as a locus.** A point `(x, y)` is as far from the focus `(0, p)` as from
the directrix, the line `y = −p`, exactly when `x² = 4py`. The distance to the focus is
`√(x² + (y − p)²)` and the distance to the line is `|y + p|`. Squaring both (they are
non-negative), `x² + (y − p)² = (y + p)²`, which simplifies to `x² = 4py`. -/
theorem parabola_focus_directrix (p x y : ℝ) :
    √(x ^ 2 + (y - p) ^ 2) = |y + p| ↔ x ^ 2 = 4 * p * y := by
  rw [← Real.sqrt_sq_eq_abs, Real.sqrt_inj (by positivity) (by positivity)]
  constructor <;> intro h <;> nlinarith [h]