Precalculus

Focus and directrix

A circle needs only a center and a radius. Is there one rule, with a single number to turn, that draws every other conic?

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Conic sections from a double conePrecalculus

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The parabola as a locusPrecalculus

Where you are 6 in Conic sections

Geometry11

Conic sections6

Conic sections from a double cone

Focus and directrix

The ellipse The hyperbola The parabola as a locus Identifying and translating conics

Why this exists

Cutting a double cone showed that the ellipse, parabola and hyperbola are one family: tip the plane a little and the curve changes kind. But that picture needs a cone in space, and each curve’s own lesson describes it a different way. The ellipse is a constant sum of distances to two foci, the hyperbola a constant difference, and the parabola is equal distances to a point and a line.

This lesson gives one rule for all three, in the plane: a point, a line, and one number. Turning that number moves through ellipse, parabola and hyperbola in order, the way tilting the plane did, and gives every conic a short polar equation with a focus at the pole.

Definition

Fix a point F, the focus, a line L that does not pass through F, the directrix, and a number e > 0, the eccentricity. The conic with this focus, directrix and eccentricity is the set of all points P in the plane with

\frac{PF}{PL} = e, \qquad \text{that is,} \qquad PF = e \cdot PL

where PL is the distance from P to the line L, measured along the perpendicular. It is a locus: every point that meets one condition.

If e < 1, the conic is an ellipse.
If e = 1, the conic is a parabola.
If e > 1, the conic is a hyperbola.

Write d for the distance from F to L. The letter e here is a ratio, not the number 2.718\ldots of exponential growth.

Intuition

Figure 1Every point P of this ellipse has \displaystyle PF = \frac{1}{2}PD, where PD is the distance from P straight across to the directrix.

In Figure 1, \displaystyle e = \frac{1}{2}: every point of the curve is half as far from F as from the directrix, so the curve stays on the focus side and closes up around F.

Now raise e. A larger e lets P wander farther from F, so the curve stretches away from the directrix. At e = 1 the point must be exactly as far from F as from L, and the curve never closes: a parabola. Past e = 1, a point on the far side of the directrix can satisfy the rule too, because crossing the line puts it farther from F than from L. That is the second branch of a hyperbola.

Figure 2The same focus and directrix with \displaystyle e = \frac{1}{2}, e = 1 and e = 2. Only e > 1 reaches past the directrix.

Going the other way, shrink e toward 0 and the ellipse grows rounder. A circle is the limit, but it is not itself one of these curves: e = 0 would demand PF = 0, which is the single point F. As e shrinks with the size held fixed, the directrix slides off without end. A circle has a center but no directrix.

Results

One ratio sorts the conics. The conic with eccentricity e is an ellipse for 0 < e < 1, a parabola for e = 1 and a hyperbola for e > 1. Every ellipse that is not a circle, every parabola and every hyperbola arises this way.

Polar form. Put the focus at the pole of Polar coordinates and the directrix at distance d from it, perpendicular or parallel to the polar axis. Then the conic is

Directrix	Polar equation
x = d	r = \dfrac{ed}{1 + e\cos\theta}
x = -d	r = \dfrac{ed}{1 - e\cos\theta}
y = d	r = \dfrac{ed}{1 + e\sin\theta}
y = -d	r = \dfrac{ed}{1 - e\sin\theta}

This is the form \displaystyle r = \frac{\ell}{1 + e\cos\theta} from converting between polar and rectangular, with \ell = ed, the value of r at right angles to the axis. To read an equation, first divide so the constant in the denominator is 1; the coefficient of \cos\theta or \sin\theta is then e.

Eccentricity from the center. For an ellipse or a hyperbola with center-to-vertex distance a and center-to-focus distance c,

e = \frac{c}{a}

and each focus has its own directrix, perpendicular to the axis through the foci, at distance \displaystyle \frac{a}{e} from the center on the same side as that focus. With center at the origin and foci (\pm c, 0), the directrices are \displaystyle x = \pm\frac{a}{e}.

Why it is true

The rectangular equation. Put F at the origin and L at x = d. Since PF and PL are never negative, PF = e \cdot PL says the same as PF^2 = e^2\,PL^2:

x^2 + y^2 = e^2(x - d)^2

Multiplying out and gathering terms,

(1 - e^2)x^2 + 2e^2dx + y^2 = e^2d^2

If e = 1, the x^2 terms cancel and y^2 = d^2 - 2dx, a parabola with its vertex at \displaystyle x = \frac{d}{2}, halfway between focus and directrix. If e \ne 1, complete the square in x:

\begin{aligned} (1 - e^2)(x - h)^2 + y^2 &= \frac{e^2d^2}{1 - e^2}, \\ h &= -\frac{e^2d}{1 - e^2} \end{aligned}

Let \displaystyle a = \frac{ed}{|1 - e^2|} and divide by the right side. For e < 1 this is \displaystyle \frac{(x - h)^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1, an ellipse with b^2 = a^2 - (ea)^2, so c = ea. For e > 1 the y^2 term turns negative: \displaystyle \frac{(x - h)^2}{a^2} - \frac{y^2}{a^2(e^2 - 1)} = 1, a hyperbola with b^2 = (ea)^2 - a^2, so again c = ea. In both, |h| = ea = c: the origin really is a focus. So \displaystyle e = \frac{c}{a}, and the distance from the center to the directrix is \displaystyle |h - d| = \frac{d}{|1 - e^2|} = \frac{a}{e}.

The polar equation. Write P = (r\cos\theta, r\sin\theta) with r \ge 0, so PF = r and PL = |d - r\cos\theta|. On the focus side of L, r = e(d - r\cos\theta), which gives r(1 + e\cos\theta) = ed. Beyond L, r = e(r\cos\theta - d), which is only possible when e\cos\theta > 1, so only when e > 1. Such a point at distance \rho in direction \theta is also the polar point (-\rho, \theta + \pi) (negative r), and \displaystyle \frac{ed}{1 + e\cos(\theta + \pi)} = \frac{ed}{1 - e\cos\theta} = -\rho. So the single equation \displaystyle r = \frac{ed}{1 + e\cos\theta} draws both branches, the far one with negative r. The other three rows of the table come the same way, with PL = |d + r\cos\theta|, |d - r\sin\theta| or |d + r\sin\theta|.

Worked examples

Example 1

Finding the eccentricity from one point

Thorgerdur knows a conic has focus (4, 0) and directrix x = 1, and passes through (4, 6). Find e, name the conic, and find where it crosses the x-axis.

Solution

The point (4, 6) is 6 from the focus and |4 - 1| = 3 from the directrix, so \displaystyle e = \frac{6}{3} = 2. Since e > 1, the conic is a hyperbola.

A point (x, 0) is on it when |x - 4| = 2|x - 1|. Either x - 4 = 2(x - 1), giving x = -2, or x - 4 = -2(x - 1), giving 3x = 6 and x = 2. The crossings are (2, 0) and (-2, 0). Check: (2, 0) is 2 from F and 1 from L; (-2, 0) is 6 from F and 3 from L. The directrix x = 1 runs between them, one vertex on each branch.

Example 2

An ellipse from focus and directrix

Sigurlaug takes the focus at the origin, the directrix x = 6, and \displaystyle e = \frac{1}{2}. Find a rectangular equation and check \displaystyle e = \frac{c}{a}.

Solution

The condition is \displaystyle \sqrt{x^2 + y^2} = \frac{1}{2}|6 - x|. Neither side is negative, so squaring loses nothing. Multiply by 4 first:

4(x^2 + y^2) = (6 - x)^2 = 36 - 12x + x^2

so 3x^2 + 12x + 4y^2 = 36. Complete the square: 3(x^2 + 4x + 4) + 4y^2 = 36 + 12, and dividing by 48,

\frac{(x + 2)^2}{16} + \frac{y^2}{12} = 1

The center is (-2, 0) with a = 4, and the focus at the origin is c = 2 from the center. So \displaystyle \frac{c}{a} = \frac{2}{4} = \frac{1}{2} = e. The directrix should sit \displaystyle \frac{a}{e} = 8 from the center: -2 + 8 = 6, as given.

Example 3

A polar equation from focus and directrix

Write the polar equation of the conic with focus at the pole, \displaystyle e = \frac{2}{3} and directrix x = 5. Find its vertices, its center and c.

Solution

The directrix is x = d with d = 5, so \displaystyle ed = \frac{10}{3} and

r = \frac{\frac{10}{3}}{1 + \frac{2}{3}\cos\theta} = \frac{10}{3 + 2\cos\theta}

after multiplying top and bottom by 3. Since e < 1, it is an ellipse. At \theta = 0, \displaystyle r = \frac{10}{5} = 2, the vertex (2, 0); at \theta = \pi, \displaystyle r = \frac{10}{1} = 10, the vertex (-10, 0). So 2a = 12, a = 6, the center is the midpoint (-4, 0), and c = 4. Check: \displaystyle \frac{c}{a} = \frac{4}{6} = \frac{2}{3}. At \displaystyle \theta = \frac{\pi}{2}, \displaystyle r = \frac{10}{3}: the point \displaystyle \left(0, \frac{10}{3}\right) is \displaystyle \frac{10}{3} from F and 5 from L, ratio \displaystyle \frac{2}{3}.

Example 4

Reading a polar equation, negative r included

Name the conic \displaystyle r = \frac{6}{2 - 3\cos\theta}, and find its eccentricity, directrix and vertices.

Solution

Divide top and bottom by 2 so the denominator starts with 1: \displaystyle r = \frac{3}{1 - \frac{3}{2}\cos\theta}. So \displaystyle e = \frac{3}{2}, a hyperbola, and ed = 3 gives d = 2. The minus sign with cosine puts the directrix at x = -2.

At \theta = \pi, \displaystyle r = \frac{6}{2 + 3} = \frac{6}{5}, the point \displaystyle \left(-\frac{6}{5}, 0\right). At \theta = 0, \displaystyle r = \frac{6}{2 - 3} = -6: negative, so the point is 6 units in the opposite direction, (-6, 0). Both vertices lie left of the focus. Check (-6, 0): it is 6 from F and 4 from x = -2, ratio \displaystyle \frac{3}{2}. The center is \displaystyle \left(-\frac{18}{5}, 0\right), so \displaystyle a = \frac{12}{5}, \displaystyle c = \frac{18}{5}, and \displaystyle \frac{c}{a} = \frac{3}{2}.

Example 5

The case e = 1

Find polar and rectangular equations for the conic with focus at the origin, directrix y = -3 and e = 1.

Solution

The directrix is y = -d with d = 3, so \displaystyle r = \frac{3}{1 - \sin\theta}. With e = 1 it is a parabola. In rectangular form, r = e \cdot PL reads \displaystyle \sqrt{x^2 + y^2} = y + 3; squaring, x^2 + y^2 = y^2 + 6y + 9, so

y = \frac{x^2}{6} - \frac{3}{2}

The vertex \displaystyle \left(0, -\frac{3}{2}\right) is halfway between focus and directrix, and (3, 0) is 3 from the focus and 3 from y = -3. The parabola gets its own full treatment in the parabola as a locus.

Example 6

Finding the directrices of an ellipse

Arnaldur draws an ellipse with vertices (\pm 10, 0) and foci (\pm 6, 0). Find e and both directrices, and check the rule at the co-vertex.

Solution

Here a = 10 and c = 6, so \displaystyle e = \frac{6}{10} = \frac{3}{5}. The directrices are \displaystyle x = \pm\frac{a}{e} = \pm 10 \cdot \frac{5}{3} = \pm\frac{50}{3}, the focus (6, 0) paired with \displaystyle x = \frac{50}{3}.

The co-vertex has b^2 = a^2 - c^2 = 100 - 36 = 64, so it is (0, 8). Its distance to (6, 0) is \displaystyle \sqrt{36 + 64} = 10, and its distance to \displaystyle x = \frac{50}{3} is \displaystyle \frac{50}{3}. The ratio is \displaystyle 10 \cdot \frac{3}{50} = \frac{3}{5}, as it must be.

Example 7

A comet’s orbit

Figure 3Example 7. A comet orbit with e = 0.9 and the Sun at the focus.

Ignoring the small pulls of the planets, a comet moving under the Sun’s gravity follows a conic with the Sun at a focus. Sigurlaug models a comet with e = 0.9 whose closest approach is 1 unit from the Sun, where one unit is Earth’s average distance from the Sun. Put the Sun at the pole and the closest point at \theta = 0. Find the orbit’s polar equation, the farthest distance, and the distance when \theta = 90^\circ.

Solution

The closest point, at \theta = 0, lies between focus and directrix, so the directrix is x = d and \displaystyle r = \frac{ed}{1 + 0.9\cos\theta}. At \theta = 0, \displaystyle \frac{ed}{1.9} = 1, so ed = 1.9:

r = \frac{1.9}{1 + 0.9\cos\theta}

Since e < 1 the orbit is an ellipse and the comet returns. The farthest point is at \theta = \pi: \displaystyle r = \frac{1.9}{0.1} = 19 units. At \theta = 90^\circ, \cos\theta = 0 and r = 1.9 units. Check: \displaystyle a = \frac{1 + 19}{2} = 10 and c = 10 - 1 = 9, so \displaystyle \frac{c}{a} = 0.9.

Boundaries

The focus must not lie on the directrix. If it does, the rule gives only degenerate conics. With F at the origin and L the y-axis, the condition is x^2 + y^2 = e^2x^2: for e < 1 only the point F, for e = 1 the line y = 0, and for e > 1 the two lines \displaystyle y = \pm\sqrt{e^2 - 1}\,x.

e must be positive, and no e gives a circle (Exercise 20).

The polar table needs the focus at the pole and a directrix perpendicular or parallel to the polar axis. Any other placement needs a shift or a rotation first. For e > 1, part of the curve comes from negative values of r; for e = 1 there is no r in the one direction the parabola opens toward; for e < 1, r is positive in every direction.

\displaystyle e = \frac{c}{a} is for ellipses and hyperbolas only. A parabola has no center, so it has no a or c; its eccentricity is 1 straight from the definition.

Where this goes next

The next lesson, the parabola as a locus, works the e = 1 case in full, with the focus and directrix anywhere. The ellipse and the hyperbola develop their two-focus definitions and standard equations, where \displaystyle e = \frac{c}{a} measures how flat or how open the curve is. Identifying a conic from its equation reads the same kinds from the squared terms.

Exercises

From the definition

A conic has focus (0, 0) and directrix x = 5 and passes through (3, 4). Find e and name the conic.
A conic has focus (2, 1) and directrix y = -3 and passes through (5, 5). Find e and name the conic.
The parabola has focus (0, 0) and directrix y = 4. Which of (4, 0), (2, 1) and (-4, 0) lie on it?
Find where the conic with focus (6, 0), directrix x = 2 and e = 3 crosses the x-axis, and name it.
Find where the conic with focus (0, 0), directrix x = -9 and \displaystyle e = \frac{1}{2} crosses the x-axis. Name it and find its center.

Equations from a focus and a directrix

Focus (0, 0), directrix x = 3, \displaystyle e = \frac{1}{2}. Find a rectangular equation in standard form.
Focus (0, 0), directrix x = 1, e = 2. Find a rectangular equation and the vertices.
Focus (0, 0), directrix y = -6, e = 1. Find a rectangular equation.
Focus (0, 2), directrix y = 8, \displaystyle e = \frac{1}{2}. Find a rectangular equation in standard form.

Polar form

In Exercises 10 to 12 the focus is at the pole. Write the polar equation and name the conic.

\displaystyle e = \frac{1}{4}, directrix x = 8
e = 3, directrix y = -1
e = 1, directrix x = -5
Name \displaystyle r = \frac{15}{5 + 3\cos\theta}, and give its eccentricity, directrix and vertices.
Name \displaystyle r = \frac{8}{2 - 4\sin\theta}, and give its eccentricity, directrix and vertices.
Convert \displaystyle r = \frac{4}{2 - \cos\theta} to a rectangular equation, and give e.

Eccentricity as c over a

An ellipse has vertices (0, \pm 13) and foci (0, \pm 5). Find e and the directrices.
A hyperbola has vertices (\pm 3, 0) and foci (\pm 5, 0). Find e and the directrices.
A comet’s closest approach is 0.5 units from the Sun and its orbit has e = 0.75. With the Sun at the pole and the closest point at \theta = 0, find the polar equation, the farthest distance, and a.

Reasoning

Explain why, when e \le 1, no point of the conic lies on the far side of the directrix.
Explain why no choice of focus, directrix and e gives a circle.
Show that scaling a conic by a factor k > 0 about its focus gives a conic with the same e, so all conics with the same e have the same shape.
Arnaldur says \displaystyle r = \frac{6}{1 + 2\cos\theta} and \displaystyle r = \frac{6}{2 + \cos\theta} are the same kind of conic, since both have a 2 and a \cos\theta. Explain what is wrong and name each.

Answers

\displaystyle e = \frac{5}{2}, a hyperbola. \displaystyle PF = \sqrt{9 + 16} = 5 and PL = |3 - 5| = 2.
\displaystyle e = \frac{5}{8}, an ellipse. \displaystyle PF = \sqrt{3^2 + 4^2} = 5 and PL = 5 - (-3) = 8.
(4, 0) and (-4, 0). Each is 4 from F and 4 from y = 4. The point (2, 1) is \displaystyle \sqrt{5} from F but 3 from the line.
(0, 0) and (3, 0), a hyperbola. From |x - 6| = 3|x - 2|: x - 6 = 3x - 6 gives x = 0, and x - 6 = -3x + 6 gives x = 3.
(9, 0) and (-3, 0), an ellipse with center (3, 0). From \displaystyle |x| = \frac{1}{2}|x + 9|: \displaystyle x = \frac{1}{2}(x + 9) gives x = 9, and \displaystyle -x = \frac{1}{2}(x + 9) gives x = -3. Then a = 6, c = 3, \displaystyle \frac{c}{a} = \frac{1}{2}.
\displaystyle \frac{(x + 1)^2}{4} + \frac{y^2}{3} = 1. From 4(x^2 + y^2) = (3 - x)^2: 3x^2 + 6x + 4y^2 = 9, and 3(x + 1)^2 + 4y^2 = 12.
3x^2 - 8x - y^2 + 4 = 0, vertices \displaystyle \left(\frac{2}{3}, 0\right) and (2, 0). From x^2 + y^2 = 4(1 - x)^2. In standard form \displaystyle \frac{(x - \frac{4}{3})^2}{\frac{4}{9}} - \frac{y^2}{\frac{4}{3}} = 1, a hyperbola with \displaystyle a = \frac{2}{3}.
x^2 = 12y + 36, that is \displaystyle y = \frac{x^2}{12} - 3. From x^2 + y^2 = (y + 6)^2.
\displaystyle \frac{x^2}{12} + \frac{y^2}{16} = 1. From 4x^2 + 4(y - 2)^2 = (8 - y)^2, the -16y terms cancel and 4x^2 + 3y^2 = 48.
\displaystyle r = \frac{8}{4 + \cos\theta}, an ellipse. Here ed = 2, so \displaystyle r = \frac{2}{1 + \frac{1}{4}\cos\theta}.
\displaystyle r = \frac{3}{1 - 3\sin\theta}, a hyperbola. Here ed = 3 and the directrix is below the pole.
\displaystyle r = \frac{5}{1 - \cos\theta}, a parabola.
An ellipse, \displaystyle e = \frac{3}{5}, directrix x = 5, vertices \displaystyle \left(\frac{15}{8}, 0\right) and \displaystyle \left(-\frac{15}{2}, 0\right). Dividing by 5 gives \displaystyle \frac{3}{1 + \frac{3}{5}\cos\theta}, so ed = 3. At \theta = 0, \displaystyle r = \frac{15}{8}; at \theta = \pi, \displaystyle r = \frac{15}{2}.
A hyperbola, e = 2, directrix y = -2, vertices (0, -4) and \displaystyle \left(0, -\frac{4}{3}\right). Dividing by 2 gives \displaystyle \frac{4}{1 - 2\sin\theta}, so ed = 4. At \displaystyle \theta = \frac{\pi}{2}, r = -4, the point (0, -4); at \displaystyle \theta = \frac{3\pi}{2}, \displaystyle r = \frac{4}{3}.
3x^2 - 8x + 4y^2 = 16, with \displaystyle e = \frac{1}{2}. From 2r - r\cos\theta = 4: 2r = x + 4, and squaring, 4x^2 + 4y^2 = x^2 + 8x + 16. Dividing the polar form by 2 shows e.
\displaystyle e = \frac{5}{13}, directrices \displaystyle y = \pm\frac{169}{5}, since \displaystyle \frac{a}{e} = 13 \cdot \frac{13}{5}.
\displaystyle e = \frac{5}{3}, directrices \displaystyle x = \pm\frac{9}{5}, since \displaystyle \frac{a}{e} = 3 \cdot \frac{3}{5}.
\displaystyle r = \frac{0.875}{1 + 0.75\cos\theta}, farthest 3.5 units, a = 2. At \theta = 0, \displaystyle \frac{ed}{1.75} = 0.5 gives ed = 0.875; at \theta = \pi, \displaystyle r = \frac{0.875}{0.25} = 3.5; \displaystyle a = \frac{0.5 + 3.5}{2}.
If P is on the far side, the segment PF crosses L, so PF is longer than the distance from P to L: PF > PL. Then \displaystyle \frac{PF}{PL} > 1 \ge e, so P is not on the conic.
For e \ge 1 the curve is not closed. For e < 1 it is an ellipse with b^2 = a^2(1 - e^2) < a^2, so its two axes have different lengths, and a circle’s do not.
Scaling about F multiplies every distance by k: PF becomes k \cdot PF, the directrix moves to distance kd, and PL becomes k \cdot PL. The ratio \displaystyle \frac{PF}{PL} is unchanged. In polar form, \displaystyle r = \frac{e(kd)}{1 + e\cos\theta} is k times the old r in every direction.
The eccentricity is read only after the constant in the denominator is 1. The first already has it: e = 2, a hyperbola. The second becomes \displaystyle \frac{3}{1 + \frac{1}{2}\cos\theta}: \displaystyle e = \frac{1}{2}, an ellipse.