---
title: Conic Sections
module: Parametric Equations and Polar Coordinates
moduleNumber: 8
lessonNumber: 3
order: 803
summary: >
  Parabolas, ellipses, and hyperbolas are the plane curves cut from a double
  cone. Each has a focus-based geometric definition and a standard Cartesian
  equation. A single number, the eccentricity, ties the three together, and
  placing a focus at the pole gives all of them one polar equation that describes
  planetary orbits.
topics: [Parametric Equations and Polar Coordinates]
sources:
  - book: Stewart
    ref: "Ch. 10 — Parametric Equations and Polar Coordinates; §10.5 Conic Sections, §10.6 Conic Sections in Polar Coordinates"
---

Slice a double cone with a plane and the edge of the cut is a **conic section**.
The angle of the plane relative to the cone's axis decides the curve: a shallow
cut gives an ellipse (a circle when horizontal), a cut parallel to the cone's
side gives a parabola, and a steep cut through both nappes gives a hyperbola.

$$
% caption: The three conics as plane sections of a double cone: a shallow cut
% gives an ellipse, a cut parallel to the slant gives a parabola, and a steep cut
% through both nappes gives a hyperbola.
\begin{tikzpicture}[>=stealth, scale=0.95, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  % double cone (hourglass) with elliptical rims
  \draw[black] (-1.3,2.4) -- (0,0) -- (1.3,2.4);
  \draw[black] (-1.3,-2.4) -- (0,0) -- (1.3,-2.4);
  \draw[black] (0,2.4) ellipse (1.3 and 0.34);
  \draw[black, dashed] (0,-2.4) ellipse (1.3 and 0.34);
  % ellipse cut (shallow)
  \draw[acc, thick] (-0.95,1.35) -- (0.95,1.75);
  \node[font=\footnotesize, anchor=west] at (1.0,1.75) {ellipse};
  % parabola cut (parallel to right slant)
  \draw[acc, thick] (-0.75,-2.15) -- (0.72,0.55);
  \node[font=\footnotesize, anchor=north] at (-0.75,-2.15) {parabola};
  % hyperbola cut (steep, both nappes)
  \draw[acc, thick] (0.62,2.5) -- (0.62,-2.5);
  \node[font=\footnotesize, anchor=south] at (0.62,2.55) {hyperbola};
  \fill[black] (0,0) circle (1.3pt);
\end{tikzpicture}
$$

The section definitions are geometric, stated in terms of distances to fixed
points and lines. Each yields a clean equation once the curve is placed
symmetrically on the axes.

## Parabolas

> **Definition (Parabola).** The set of points equidistant from a fixed point
> $F$ (the **focus**) and a fixed line (the **directrix**). The point halfway
> between them is the **vertex**; the line through the focus perpendicular to the
> directrix is the **axis**.

Place the vertex at the origin with focus $F(0, p)$ and directrix $y = -p$. A
point $P(x, y)$ lies on the parabola when its distance to the focus equals its
distance to the directrix.

$$
% caption: A parabola with focus $F=(0,p)$ and directrix $y=-p$: every point $P$
% is equidistant from the focus and the directrix, so $PF = PM$ where $M$ is the
% foot of the perpendicular to the directrix.
\begin{tikzpicture}[>=stealth, scale=0.95, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[very thick] plot[domain=-2.1:2.1, samples=60] (\x, {0.5*\x*\x});
  \draw[black] (-2.4,-0.5) -- (2.4,-0.5) node[right, font=\footnotesize] {directrix y=-p};
  \fill[black] (0,0.5) circle (2pt) node[right, font=\footnotesize] {$F$};
  \coordinate (P) at (1.55,1.20);
  \fill[acc] (P) circle (2.4pt) node[above right, font=\footnotesize] {$P$};
  \coordinate (M) at (1.55,-0.5);
  \fill[black] (M) circle (1.6pt) node[below, font=\footnotesize] {$M$};
  \draw[acc, dashed] (P) -- (M) node[midway, right, font=\footnotesize] {$PM$};
  \draw[acc, dashed] (P) -- (0,0.5) node[midway, above left, font=\footnotesize] {$PF$};
  \fill[black] (0,0) circle (1.4pt);
  \node[font=\footnotesize, anchor=north east] at (0,0) {vertex};
\end{tikzpicture}
$$

Equating $\sqrt{x^2 + (y-p)^2} = |y + p|$, squaring, and cancelling gives

$$
x^2 + (y-p)^2 = (y+p)^2 \;\Longrightarrow\; x^2 = 4py.
$$

> **Theorem (Standard parabola).** The parabola with focus $(0, p)$ and directrix
> $y = -p$ is $x^2 = 4py$, opening up if $p > 0$ and down if $p < 0$.
> Interchanging $x$ and $y$ gives $y^2 = 4px$, with focus $(p, 0)$, directrix
> $x = -p$, opening right or left.

> **Worked example.** Find the focus and directrix of $y^2 + 10x = 0$.
>
> Rewrite as $y^2 = -10x$ and compare with $y^2 = 4px$: then $4p = -10$, so
> $p = -\tfrac52$. The parabola opens left, with focus $(-\tfrac52, 0)$ and
> directrix $x = \tfrac52$.

## Ellipses

> **Definition (Ellipse).** The set of points the sum of whose distances from two
> fixed points $F_1$, $F_2$ (the **foci**) is a constant $2a$.

Put the foci at $(\pm c, 0)$. The condition $|PF_1| + |PF_2| = 2a$, squared twice
to clear the radicals, reduces to $(a^2 - c^2)x^2 + a^2 y^2 = a^2(a^2 - c^2)$.
Since $c < a$, set $b^2 = a^2 - c^2 > 0$ and divide through.

$$
% caption: An ellipse with foci $F_1, F_2$ at $(\pm c, 0)$: for every point $P$
% the focal radii satisfy $|PF_1| + |PF_2| = 2a$; the semi-axes obey
% $b^2 = a^2 - c^2$.
\begin{tikzpicture}[>=stealth, scale=1.0, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[very thick] (0,0) ellipse (3 and 2);
  \draw[black] (-3,0) -- (3,0);
  \draw[black] (0,-2) -- (0,2);
  \coordinate (F1) at (-2.236,0);
  \coordinate (F2) at (2.236,0);
  \fill[black] (F1) circle (2pt) node[below, font=\footnotesize] {$F_1$};
  \fill[black] (F2) circle (2pt) node[below, font=\footnotesize] {$F_2$};
  \coordinate (P) at (1.03,1.71);
  \fill[acc] (P) circle (2.4pt) node[above right, font=\footnotesize] {$P$};
  \draw[acc] (F1) -- (P) node[midway, above left, font=\footnotesize] {$r_1$};
  \draw[acc] (F2) -- (P) node[midway, above right, font=\footnotesize] {$r_2$};
  \node[font=\footnotesize, anchor=west] at (3.12,0) {$a$};
  \node[font=\footnotesize, anchor=south] at (0,2.14) {$b$};
  \node[font=\footnotesize, anchor=north] at (-1.1,-0.34) {$c$};
  \draw[black, |-|] (0,-0.28) -- (-2.236,-0.28);
\end{tikzpicture}
$$

> **Theorem (Standard ellipse).** With $a > b > 0$ and $c^2 = a^2 - b^2$,
> $$
> \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
> $$
> has foci $(\pm c, 0)$, vertices $(\pm a, 0)$, major axis along $x$, and minor
> axis along $y$. Foci on the $y$-axis swap the roles: $\dfrac{x^2}{b^2} +
> \dfrac{y^2}{a^2} = 1$ with foci $(0, \pm c)$.

> **Worked example.** Locate the vertices and foci of $9x^2 + 16y^2 = 144$.
>
> Divide by $144$:
>
> $$
> \frac{x^2}{16} + \frac{y^2}{9} = 1.
> $$
>
> Then $a^2 = 16$ and $b^2 = 9$, so $a = 4$, $b = 3$, and $c^2 = a^2 - b^2 = 7$. The
> major axis lies along $x$, the vertices are $(\pm 4, 0)$, and the foci are
> $(\pm\sqrt7, 0)$.

> **Worked example.** Find the equation of the ellipse with foci $(0, \pm 2)$ and
> vertices $(0, \pm 3)$.
>
> The foci and vertices lie on the $y$-axis, so the major axis is vertical:
> $c = 2$, $a = 3$, and $b^2 = a^2 - c^2 = 9 - 4 = 5$. With the larger denominator
> under $y^2$,
>
> $$
> \frac{x^2}{5} + \frac{y^2}{9} = 1.
> $$

If the foci coincide ($c = 0$) then $a = b$ and the ellipse is a circle.

Ellipses and parabolas share a **reflection property**, a consequence of the equal
angles a tangent makes with the focal radii. A ray leaving one focus of an ellipse
reflects off the curve straight to the other focus; lithotripsy exploits this by
placing a kidney stone at one focus and a shock-wave source at the other. A
parabola is the limiting case with one focus sent to infinity: rays from the focus
reflect into a parallel beam, which is why headlamp reflectors and satellite dishes
have parabolic cross-sections.

$$
% caption: Reflection property of the ellipse: a ray from focus $F_1$ meets the
% curve at $P$ and reflects to $F_2$, since the tangent at $P$ makes equal angles
% with the two focal radii.
\begin{tikzpicture}[>=stealth, scale=1.0, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[very thick] (0,0) ellipse (3 and 2);
  \coordinate (F1) at (-2.236,0);
  \coordinate (F2) at (2.236,0);
  \fill[black] (F1) circle (2pt) node[below, font=\footnotesize] {$F_1$};
  \fill[black] (F2) circle (2pt) node[below, font=\footnotesize] {$F_2$};
  \coordinate (P) at (1.03,1.71);
  \fill[acc] (P) circle (2.2pt);
  \draw[acc, ->] (F1) -- (P);
  \draw[acc, ->] (P) -- (F2);
  \draw[black, dashed] ($(P)+(-1.3,0.34)$) -- ($(P)+(1.3,-0.34)$);
  \node[font=\footnotesize, anchor=south west] at (P) {$P$};
\end{tikzpicture}
$$

## Hyperbolas

> **Definition (Hyperbola).** The set of points the _difference_ of whose
> distances from two foci $F_1$, $F_2$ is a constant $\pm 2a$.

The derivation mirrors the ellipse, but now $c > a$, so $b^2 = c^2 - a^2 > 0$. The
curve has two branches and no $y$-intercept, and each branch approaches a pair of
straight **asymptotes**.

$$
% caption: A hyperbola $x^2/a^2 - y^2/b^2 = 1$ with foci at $(\pm c, 0)$,
% $c^2 = a^2 + b^2$: the two branches open along the $x$-axis and hug the
% asymptotes $y = \pm(b/a)x$.
\begin{tikzpicture}[>=stealth, scale=0.85, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (-3.4,0) -- (3.4,0) node[right, font=\footnotesize] {$x$};
  \draw[black] (0,-3.2) -- (0,3.2) node[above, font=\footnotesize] {$y$};
  \draw[black, dashed] (-3.0,-3.0) -- (3.0,3.0);
  \draw[black, dashed] (-3.0,3.0) -- (3.0,-3.0);
  \node[black, font=\footnotesize, anchor=south west] at (2.5,2.5) {asymptote};
  \draw[acc, very thick] plot[domain=1.5:3.0, samples=40] (\x, {1.5*sqrt((\x/1.5)^2 - 1)});
  \draw[acc, very thick] plot[domain=1.5:3.0, samples=40] (\x, {-1.5*sqrt((\x/1.5)^2 - 1)});
  \draw[acc, very thick] plot[domain=1.5:3.0, samples=40] (-\x, {1.5*sqrt((\x/1.5)^2 - 1)});
  \draw[acc, very thick] plot[domain=1.5:3.0, samples=40] (-\x, {-1.5*sqrt((\x/1.5)^2 - 1)});
  \fill[black] (2.12,0) circle (2pt) node[below right, font=\footnotesize] {$F_2$};
  \fill[black] (-2.12,0) circle (2pt) node[below left, font=\footnotesize] {$F_1$};
  \fill (1.5,0) circle (1.6pt);
  \node[font=\footnotesize, anchor=north] at (1.5,-0.1) {$a$};
\end{tikzpicture}
$$

> **Theorem (Standard hyperbola).** With $c^2 = a^2 + b^2$,
> $$
> \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
> $$
> has foci $(\pm c, 0)$, vertices $(\pm a, 0)$, and asymptotes
> $y = \pm\dfrac{b}{a}x$. The conjugate form $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2}
> = 1$ opens along the $y$-axis.

> **Worked example.** Find the foci and asymptotes of $9x^2 - 16y^2 = 144$.
>
> Divide by $144$:
>
> $$
> \frac{x^2}{16} - \frac{y^2}{9} = 1,
> $$
>
> so $a = 4$, $b = 3$, and $c^2 = a^2 + b^2 = 25$. The foci are $(\pm 5, 0)$ and the
> asymptotes are $y = \pm\tfrac34 x$.

**Shifted conics.** Replacing $x$ by $x - h$ and $y$ by $y - k$ moves the center to
$(h, k)$; completing the square recovers the standard form from a general
second-degree equation.

> **Worked example.** Identify the conic $9x^2 - 4y^2 - 72x + 8y + 176 = 0$.
>
> Group the $x$- and $y$-terms and factor out the leading coefficients:
>
> $$
> 9(x^2 - 8x) - 4(y^2 - 2y) = -176.
> $$
>
> Complete each square, adding $9\cdot 16$ and subtracting $4\cdot 1$ on the left,
> and the same amounts on the right:
>
> $$
> 9(x - 4)^2 - 4(y - 1)^2 = -176 + 144 - 4 = -36.
> $$
>
> Dividing by $-36$ gives standard hyperbola form:
>
> $$
> \frac{(y - 1)^2}{9} - \frac{(x - 4)^2}{4} = 1,
> $$
>
> a hyperbola centered at $(4, 1)$ opening vertically, with $a = 3$, $b = 2$, and
> $c^2 = 13$, so its foci are $(4,\, 1 \pm \sqrt{13})$.

| Conic | Focal condition | Standard equation | Key relation |
| --- | --- | --- | --- |
| Parabola | $\lvert PF\rvert = \lvert P\ell\rvert$ | $x^2 = 4py$ | vertex midway to directrix |
| Ellipse | $\lvert PF_1\rvert + \lvert PF_2\rvert = 2a$ | $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ | $b^2 = a^2 - c^2$ |
| Hyperbola | $\bigl\lvert\, \lvert PF_1\rvert - \lvert PF_2\rvert \,\bigr\rvert = 2a$ | $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ | $b^2 = c^2 - a^2$ |

## Eccentricity

The focus-and-directrix definition of the parabola generalizes to all conics by
allowing the distance ratio to differ from $1$.

> **Theorem (Focus-directrix / eccentricity).** Fix a focus $F$, a directrix
> $\ell$, and a positive number $e$ (the **eccentricity**). The set of points $P$
> with
> $$
> \frac{\lvert PF\rvert}{\lvert P\ell\rvert} = e
> $$
> is an ellipse if $e < 1$, a parabola if $e = 1$, and a hyperbola if $e > 1$.

$$
% caption: The focus-directrix ratio $|PF|/|P\ell| = e$ defines every conic at
% once; the value of $e$ selects the type, with $e<1$ an ellipse, $e=1$ a
% parabola, and $e>1$ a hyperbola.
\begin{tikzpicture}[>=stealth, scale=1.0, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, very thick] (2.4,-1.6) -- (2.4,1.9) node[above, font=\footnotesize] {directrix};
  \fill[black] (0,0) circle (2pt) node[below left, font=\footnotesize] {$F$};
  \draw[thick] (0,0) ellipse (1.5 and 1.15);
  \coordinate (P) at (0.6,1.05);
  \fill[acc] (P) circle (2.4pt) node[above, font=\footnotesize] {$P$};
  \coordinate (M) at (2.4,1.05);
  \fill[black] (M) circle (1.6pt) node[right, font=\footnotesize] {$M$};
  \draw[acc, dashed] (P) -- (0,0) node[midway, above left, font=\footnotesize] {$PF$};
  \draw[acc, dashed] (P) -- (M) node[midway, above, font=\footnotesize] {$PM$};
\end{tikzpicture}
$$

For an ellipse $e = c/a < 1$, and for a hyperbola $e = c/a > 1$. Eccentricity is a
shape number, independent of size: $e$ near $0$ is nearly circular, $e$ near $1$
is elongated, and $e$ growing past $1$ opens the branches of a hyperbola wider.

| Eccentricity | Conic |
| --- | --- |
| $e = 0$ | circle |
| $0 < e < 1$ | ellipse |
| $e = 1$ | parabola |
| $e > 1$ | hyperbola |

## The polar equation of a conic

Placing the focus at the pole turns the ratio definition into one compact polar
equation. With directrix $x = d$ to the right of the focus, $|PF| = r$ and
$|P\ell| = d - r\cos\theta$, so $r = e(d - r\cos\theta)$. Solving for $r$:[^ratio]

$$
r = \frac{ed}{1 + e\cos\theta}.
$$

> **Theorem (Polar conic).** A conic with a focus at the pole and eccentricity
> $e$ has an equation of the form
> $$
> r = \frac{ed}{1 \pm e\cos\theta} \qquad\text{or}\qquad r = \frac{ed}{1 \pm e\sin\theta},
> $$
> the sign and the choice of $\cos$/$\sin$ fixing which side the directrix lies
> on. It is an ellipse if $e < 1$, a parabola if $e = 1$, a hyperbola if $e > 1$.

The four sign-and-function combinations correspond to the four directrix
placements relative to the focus:

- $r = \dfrac{ed}{1 + e\cos\theta}$: directrix vertical, to the right of the focus.
- $r = \dfrac{ed}{1 - e\cos\theta}$: directrix vertical, to the left.
- $r = \dfrac{ed}{1 + e\sin\theta}$: directrix horizontal, above the focus.
- $r = \dfrac{ed}{1 - e\sin\theta}$: directrix horizontal, below.

> **Worked example.** Find a polar equation for the parabola with focus at the pole
> and directrix $y = -6$.
>
> A parabola has $e = 1$, and a horizontal directrix at distance $d = 6$ below the
> pole selects the $-\sin\theta$ form:
>
> $$
> r = \frac{ed}{1 - e\sin\theta} = \frac{6}{1 - \sin\theta}.
> $$

> **Worked example.** For $r = \dfrac{10}{3 - 2\cos\theta}$, find the eccentricity,
> identify the conic, and locate its directrix and vertices.
>
> Divide numerator and denominator by $3$ to reach standard form:
>
> $$
> r = \frac{10/3}{1 - \tfrac23\cos\theta}.
> $$
>
> Then $e = \tfrac23 < 1$, so the curve is an ellipse. From $ed = \tfrac{10}{3}$
> with $e = \tfrac23$, the directrix is at distance $d = 5$. The vertices lie on the
> axis at $\theta = 0$ and $\theta = \pi$, where $r = \dfrac{10}{3 - 2} = 10$ and
> $r = \dfrac{10}{3 + 2} = 2$.

## Orbits

Kepler's first law places each planet on an ellipse with the sun at one focus.
Writing the polar equation in terms of the semimajor axis $a$ and eccentricity
$e$ (using $ed = a(1 - e^2)$) gives the standard orbital form.

$$
r = \frac{a(1 - e^2)}{1 + e\cos\theta}.
$$

The extreme distances occur at the vertices. At $\theta = 0$ the planet is at
**perihelion** (closest), and at $\theta = \pi$ at **aphelion** (farthest):

$$
r_{\min} = a(1 - e), \qquad r_{\max} = a(1 + e).
$$

For Earth, $e \approx 0.017$ and $2a \approx 2.99 \times 10^8$ km, so
$a \approx 1.495 \times 10^8$ km. The orbit is
$r \approx \dfrac{1.49 \times 10^8}{1 + 0.017\cos\theta}$ km, with perihelion
about $1.47 \times 10^8$ km and aphelion about $1.52 \times 10^8$ km. A small
eccentricity keeps the orbit close to a circle, and the same equation, with $e$
pushed past $1$, describes the unbound hyperbolic path of a comet that swings
once past the sun and never returns.

[^ratio]: Stewart, §10.6 — the polar conic equation follows from the
focus-directrix ratio $|PF| = e\,|P\ell|$ with the focus at the pole and the
directrix perpendicular to the polar axis.