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Precalculus: An Investigation of Functions, by David Lippman and Melonie Rasmussen, licensed under CC BY-SA 4.0. Read the original free This shelf’s study material is shared under the same license (CC BY-SA 4.0).

Chapter 13 · pp. 589–650

Chapter 9: Conics

A conic section is a curve made when a plane slices a cone. This chapter studies the ellipse, hyperbola and parabola. It covers their definitions, their equations (including shifted ones), and uses such as bridges, planetary orbits, ship navigation and reflectors. It also covers solving non-linear systems and describing conics in polar form with eccentricity.

Conic sections and the ellipse

Imagine a double cone cut by a flat plane. The edge of the cut is a conic section. If the plane is perpendicular to the cone's axis, the cut is a circle. If the plane is tilted a little, the cut is an oval called an ellipse. Other tilts give parabolas and hyperbolas (p. 589).

An ellipse can be drawn with two thumbtacks, a loop of string and a pencil. The tacks are the two fixed points, called the foci. The string length never changes, so the pencil traces every point whose distances to the two foci add to the same constant. In symbols, for a point $Q$ on the ellipse, $d(Q,F_1)+d(Q,F_2)=k$ (p. 590).

Vocabulary of an ellipse

The major axis is the line through the two foci. The points where it meets the ellipse are the vertices. The distance between the vertices is the major axis length. The center is the midpoint of the vertices, which is also the midpoint of the foci.
The minor axis passes through the center at right angles to the major axis. It meets the ellipse at the minor axis endpoints, also called co-vertices. The distance between them is the minor axis length. (p. 590)

Which axis is major depends on how the ellipse is turned. Foci on the x-axis mean a horizontal major axis. Foci on the y-axis mean a vertical one. The major axis is always the longer axis, and the vertices always lie on it (p. 590).

The constant sum $k$ equals $2a$ , the major axis length. To see why, take the vertex $(a,0)$ . It is $a-c$ from one focus and $a+c$ from the other. Those distances add to $2a$ (p. 591).

Ellipses centered at the origin

When the major axis is horizontal and $a>b>0$ , the equation is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$

The vertices are $(\pm a,0)$ and the co-vertices are $(0,\pm b)$ . For example, $\frac{x^2}{25}+\frac{y^2}{9}=1$ has vertices $(\pm5,0)$ and co-vertices $(0,\pm3)$ (p. 592).

When the major axis is vertical, the larger denominator goes under $y$ :

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1.$

Now the vertices are $(0,\pm a)$ and the co-vertices are $(\pm b,0)$ . For $x^2+\frac{y^2}{9}=1$ , we have $a=3$ and $b=1$ (p. 592).

Reading features from an equation. Divide so the right side is 1, compare the denominators, then take square roots. For $x^2+9y^2=9$ , dividing by 9 gives $\frac{x^2}{9}+\frac{y^2}{1}=1$ . The $x$ denominator is larger, so the major axis is horizontal, with $a=3$ and $b=1$ . The major axis length is $2a=6$ and the minor axis length is $2b=2$ (p. 593).

Writing an equation from axis lengths or a graph. Halve each axis length to get $a$ and $b$ , and put $a^2$ under the variable of the major axis. A horizontal major axis of length 28 and a minor axis of length 16 give $a=14$ and $b=8$ . The equation is $\frac{x^2}{196}+\frac{y^2}{64}=1$ . On a graph, read the vertices and co-vertices to get the same two numbers (p. 594).

Ellipses centered at $(h,k)$

Shifting the ellipse $h$ units right and $k$ units up means replacing $x$ with $x-h$ and $y$ with $y-k$ :

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\ \text{(horizontal)},\qquad \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1\ \text{(vertical)}.$

Horizontal ellipses have vertices $(h\pm a,k)$ . Vertical ones have vertices $(h,k\pm a)$ . The co-vertices use $\pm b$ in the other direction (p. 595).

Standard form from a general equation. Group the $x$ terms and the $y$ terms, complete the square for each, and divide so the right side is 1. Take $x^2+2x+4y^2-24y=-32$ . Adding 1 completes the $x$ square. For the $y$ terms, factor out 4 first: $4(y^2-6y)$ . Completing the square needs 9 inside, which adds $4\cdot9=36$ to the equation. The result is $(x+1)^2+4(y-3)^2=4$ . Dividing by 4 gives $\frac{(x+1)^2}{4}+\frac{(y-3)^2}{1}=1$ . The center is $(-1,3)$ , $a=2$ and $b=1$ (p. 596).

Equation from a center, a vertex and another point. Suppose the center is $(-2,1)$ and a vertex is $(-2,4)$ . The $x$ -values match, so the major axis is vertical, and $a=4-1=3$ . Substitute the point $(0,1)$ into $\frac{(x+2)^2}{b^2}+\frac{(y-1)^2}{9}=1$ . This gives $\frac{4}{b^2}=1$ , so $b=2$ (p. 597).

Bridge arch. A half-ellipse arch over a 50 ft river that rises 20 ft is modeled by placing the center at the middle of the river, with the river as the major axis. Then $a=25$ , $b=20$ , and $\frac{x^2}{625}+\frac{y^2}{400}=1$ . At $x=15$ , $y^2=400(1-\frac{225}{625})=256$ , so the arch is 16 ft high there (p. 598). If the roadway is 4 ft above the top of the arch, it is 24 ft above the river. The gap between road and arch at that spot is $24-16=8$ ft.

Foci of an ellipse

The distance $c$ from the center to each focus satisfies

$b^2=a^2-c^2,\qquad\text{so}\qquad c^2=a^2-b^2.$

The foci lie inside the ellipse, so $c<a$ (p. 599). This looks like the Pythagorean Theorem, but it is not that theorem, and you cannot find $c$ by adding $a^2$ and $b^2$ . Adding the squares is the pattern for hyperbolas, whose foci lie beyond the vertices (pp. 599, 607).

Whispering chamber. The elliptical hall is 96 ft long and 46 ft wide, so $a=48$ and $b=23$ . Then $c=\sqrt{48^2-23^2}=\sqrt{1775}\approx42$ ft. Two people at the foci stand about $2(42)=84$ ft apart, each about $48-42=6$ ft from the end wall (p. 599).

Foci when the center is not the origin. Find the center, compute $c=\sqrt{a^2-b^2}$ , and move $c$ units from the center along the major axis. In $\frac{(x-2)^2}{4}+\frac{(y+3)^2}{29}=1$ , the larger denominator is under $y$ , so the ellipse is vertical. Here $a^2=29$ and $b^2=4$ , so $c=\sqrt{25}=5$ . The center is $(2,-3)$ , so the foci are $(2,-3\pm5)$ , which are $(2,-8)$ and $(2,2)$ (p. 599).

Equation from the foci and the major axis length. Follow these steps:

Take the center as the midpoint of the foci.
Take $a$ as half the major axis length.
Take $c$ as half the distance between the foci.
Compute $b^2=a^2-c^2$ .

Foci $(-1,4)$ and $(3,4)$ give center $(1,4)$ , a horizontal ellipse, and $c=2$ . A major axis of 10 gives $a=5$ . So $b^2=25-4=21$ , and the equation is $\frac{(x-1)^2}{25}+\frac{(y-4)^2}{21}=1$ (p. 600).

Planetary orbits

Kepler's first law says each planet moves in an ellipse with the sun at one focus. The orbits are only approximately elliptical, because bodies tug on one another. The aphelion is the planet's greatest distance from the sun, and the perihelion is its smallest. Their sum is the major axis length (p. 600).

For Mercury, the aphelion is 35.98 million miles and the perihelion is 28.58 million miles. Then $2a=64.56$ , so $a=32.28$ . The perihelion is the distance from the sun (a focus) to the nearest vertex, so it equals $a-c$ . This gives $c=32.28-28.58=3.7$ . Then $b^2=32.28^2-3.7^2\approx1028.3$ , and with the center at the origin the orbit is

$\frac{x^2}{1041.98}+\frac{y^2}{1028.31}=1.$

The values of $a^2$ and $b^2$ are nearly equal, so Mercury's orbit is close to a circle (p. 601).

Defining a hyperbola

A hyperbola is the next conic after the ellipse. It comes from slicing a double cone with a plane steep enough to cut both halves (pp. 607–608). The definition is a close cousin of the ellipse rule. For an ellipse, the distances to two foci add to a constant. For a hyperbola, the absolute value of the difference of the distances to the foci $F_1$ and $F_2$ is a constant $k$ :

$|d(Q,F_1)-d(Q,F_2)|=k$

(pp. 607–608)

Vocabulary to know (p. 608):

The transverse axis is the line through the foci.
The vertices are the points where the curve crosses the transverse axis.
The center is the midpoint of the vertices, which is also the midpoint of the foci.
The conjugate axis is the other axis of symmetry through the center.
The two separate pieces of the curve are branches, and the curve has two asymptotes.

To find where the constant comes from, put a vertex $(a,0)$ into the definition with foci $(\pm c,0)$ . The distances to the foci are $c-a$ and $c+a$ , and they differ by $2a$ . So $k=2a$ , the length of the transverse axis (p. 609). Use this whenever a problem gives you a difference in distances directly.

Standard equations centered at the origin

Setting up the distance condition and simplifying (with $b^2=c^2-a^2$ ) gives the standard forms (pp. 609–611):

Opens horizontally: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , with vertices $(\pm a,0)$ .
Opens vertically: $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ , with vertices $(0,\pm a)$ .

The opening direction depends on which squared term is positive. It does not depend on which denominator is bigger. In both forms, $a^2$ sits under the positive term.

The asymptotes are $y=\pm\frac{b}{a}x$ for a horizontal hyperbola and $y=\pm\frac{a}{b}x$ for a vertical one (pp. 610–611). The reasoning: solve the horizontal form for $y$ to get $y=\pm\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}$ . When $|x|$ is very large, $\frac{a^2}{x^2}$ is nearly 0, so the root is nearly 1. The branches therefore hug the lines $y=\pm\frac{b}{a}x$ (p. 610).

Sketching a hyperbola

Put the equation in standard form and read $a$ and $b$ .
Draw the central rectangle. Its sides are parallel to the axes and pass through the vertices and co-vertices.
Draw the rectangle's diagonals, extended. These are the asymptotes.
Start each branch at a vertex and curve it outward toward the asymptotes.

Example: $y^2-4x^2=4$ . Divide by 4 to get $\frac{y^2}{4}-x^2=1$ . The $y^2$ term is positive, so the hyperbola opens vertically. Here $a=2$ and $b=1$ . The vertices are $(0,\pm2)$ , the transverse axis has length 4, and the rectangle has corners $(\pm1,\pm2)$ . The asymptotes are $y=\pm2x$ (pp. 608, 611–612).

Writing equations at the origin

From vertices and an asymptote. Vertices $(\pm6,0)$ mean a horizontal hyperbola with $a=6$ . The asymptote $y=\frac{4}{3}x$ has slope $\frac{b}{a}=\frac{4}{3}$ , so $b=8$ . The equation is $\frac{x^2}{36}-\frac{y^2}{64}=1$ . For a vertical hyperbola, the slope is $\frac{a}{b}$ instead, so be careful about which ratio to use (p. 613).

From vertices and a point. Vertices $(0,\pm9)$ give a vertical form with $a=9$ . Substitute the point $(8,15)$ into $\frac{y^2}{81}-\frac{x^2}{b^2}=1$ to get $\frac{225}{81}-\frac{64}{b^2}=1$ . Then $\frac{64}{b^2}=\frac{16}{9}$ , so $b^2=36$ . The equation is $\frac{y^2}{81}-\frac{x^2}{36}=1$ (p. 613).

Hyperbolas centered at $(h,k)$

Shifting the center to $(h,k)$ gives (p. 614):

Horizontal: $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ . The vertices are $(h\pm a,k)$ , and the asymptotes are $y-k=\pm\frac{b}{a}(x-h)$ .
Vertical: $\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$ . The vertices are $(h,k\pm a)$ , and the asymptotes are $y-k=\pm\frac{a}{b}(x-h)$ .

The rectangle corners are $(h\pm a,k\pm b)$ in the horizontal case.

Reading a graph. If the asymptotes cross at $(2,3)$ and the vertices are $(2,2)$ and $(2,4)$ , the hyperbola is vertical with $a=1$ . If the rectangle reaches from $x=-1$ to $x=5$ , then $b=3$ . The equation is $\frac{(y-3)^2}{1}-\frac{(x-2)^2}{9}=1$ (p. 615).

General form to standard form. Group the $x$ terms and $y$ terms, and complete the square for each. Take care with the minus sign in front of the $y$ part, because the number you add inside the parentheses gets multiplied by a negative. For $9x^2+18x-4y^2+16y=43$ :

$9(x+1)^2-4(y-2)^2=36 \quad\Rightarrow\quad \frac{(x+1)^2}{4}-\frac{(y-2)^2}{9}=1$

The center is $(-1,2)$ , with $a=2$ and $b=3$ . It opens horizontally, with vertices $(-3,2)$ and $(1,2)$ (pp. 615–616).

Vertices and an asymptote, center not at the origin. Vertices $(-2,-5)$ and $(-2,7)$ share an $x$ -value, so the hyperbola is vertical. The center is their midpoint, $(-2,1)$ , and $a=6$ . The asymptote slope $\frac{3}{2}$ equals $\frac{a}{b}$ , so $b=4$ . The equation is $\frac{(y-1)^2}{36}-\frac{(x+2)^2}{16}=1$ (pp. 616–617).

Foci of a hyperbola

The distance $c$ from the center to each focus satisfies

$b^2=c^2-a^2,\quad\text{so}\quad c^2=a^2+b^2.$

This is not the Pythagorean Theorem, and it is not the ellipse rule $b^2=a^2-c^2$ either. The difference makes sense geometrically. Ellipse foci lie inside the curve, before the vertices, so $c<a$ . Hyperbola foci lie beyond the vertices, so $c>a$ (p. 617).

Example: for $\frac{(y+1)^2}{4}-\frac{(x-3)^2}{5}=1$ , the center is $(3,-1)$ and the hyperbola is vertical. Then $c=\sqrt{4+5}=3$ , and the foci are $(3,-1\pm3)$ , which are $(3,-4)$ and $(3,2)$ (p. 618).

From foci and vertices. Foci $(5,-8)$ and $(-3,-8)$ and vertices $(4,-8)$ and $(-2,-8)$ give center $(1,-8)$ . The distance to a vertex is $a=3$ , and the distance to a focus is $c=4$ . So $b^2=16-9=7$ , and the equation is $\frac{(x-1)^2}{9}-\frac{(y+8)^2}{7}=1$ (p. 618).

Application: LORAN

Before GPS, LORAN located ships using two radio stations that sent signals at the same moment. The ship measured the difference in arrival times and converted it to a difference in distance. That put the ship on a hyperbola with the two stations as foci. A second pair of stations gave a second hyperbola, and the ship lay where the two hyperbolas crossed (p. 619).

Example: stations 150 km apart, signals travel at 300,000 km/s, and one signal arrives 0.0003 s before the other. The distance difference is $300{,}000\times0.0003=90$ km. Since this equals $2a$ , $a=45$ . Half the station distance gives $c=75$ . Then $b^2=75^2-45^2=3600$ . With the center at the origin and stations at $(\pm75,0)$ , the hyperbola is

$\frac{x^2}{2025}-\frac{y^2}{3600}=1$

(p. 619).

The Parabola as a Conic

A parabola is the curve you get when a plane cuts a cone while running parallel to the cone's side. It can also be defined by distances. Pick a fixed point, the focus, and a fixed line, the directrix. The parabola is every point that is the same distance from the focus as from the directrix (p. 627).

Vocabulary:

The axis of symmetry runs through the focus and is perpendicular to the directrix.
The vertex is where the parabola crosses its axis. It sits exactly halfway between the focus and the directrix.
The focal length $p$ is the distance from the vertex to the focus (p. 627).

With the vertex at the origin and the focus at $(0,p)$ , the directrix is $y=-p$ (p. 627).

Equations with Vertex at the Origin

To get the equation, let $(x,y)$ be any point on the parabola. Set its distance to the focus equal to its distance to the directrix:

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

Square both sides and expand. The $y^2$ and $p^2$ terms cancel, leaving

$x^2=4py$

(p. 628). Dividing by $4p$ gives $y=\frac{1}{4p}x^2$ . So this is just $y=x^2$ stretched or compressed vertically by the factor $\frac{1}{4p}$ (p. 628).

A parabola with a horizontal axis has focus $(p,0)$ and a vertical directrix. Its equation is $y^2=4px$ (p. 628).

The sign of $p$ tells you which way the parabola opens:

For $x^2=4py$ : $p>0$ opens up, $p<0$ opens down.
For $y^2=4px$ : $p>0$ opens right, $p<0$ opens left (p. 628).

Example. A parabola has vertex $(0,0)$ and focus $(0,-2)$ . The focus is on the y-axis, so the axis is vertical and the form is $x^2=4py$ . Here $p=-2$ , so $x^2=4(-2)y$ , or $x^2=-8y$ . It opens down (p. 628).

Vertex at $(h,k)$

Shifting the parabola $h$ units right and $k$ units up gives two standard forms (p. 629):

	Vertical axis	Horizontal axis
Equation	$(x-h)^2=4p(y-k)$	$(y-k)^2=4p(x-h)$
Axis of symmetry	$x=h$	$y=k$
Focus	$(h,k+p)$	$(h+p,k)$
Directrix	$y=k-p$	$x=h-p$

Finding the focus and directrix from a quadratic. Isolate the squared factor so the equation matches the standard form. For $y=8(x-1)^2+2$ :

Subtract 2: $y-2=8(x-1)^2$ .
Divide by 8: $(x-1)^2=\frac18(y-2)$ .
Match $4p=\frac18$ , so $p=\frac1{32}$ .

The vertex is $(1,2)$ and the axis is $x=1$ . The focus is $\left(1,\,2+\frac1{32}\right)$ , which is $\left(1,2\frac{1}{32}\right)$ . The directrix is $y=2-\frac1{32}=1\frac{31}{32}$ (p. 630).

Writing an equation from the vertex and focus. Say the vertex is $(1,5)$ and the focus is $(3,5)$ . Both points lie on $y=5$ , so that is the axis and the parabola is horizontal. Then $h=1$ , $k=5$ , and $p=3-1=2$ . Substitute into $(y-k)^2=4p(x-h)$ :

$(y-5)^2=8(x-1)$

(pp. 630–631). Keep the sign of $p$ . If the focus were below a vertex of $(-2,3)$ at $(-2,2)$ , then $p=2-3=-1$ and the parabola would open down.

Reflection and Applications

A parabola has a reflection property. A ray leaving the focus bounces off the curve and travels parallel to the axis of symmetry. Run in reverse, parallel incoming rays all bounce to the focus (p. 631). Flashlight and headlight reflectors use the first direction to make a straight beam. Satellite dishes, radio telescopes, and solar cookers use the second direction to collect energy at one point (p. 631).

Example: solar cooker. A dish is 16 in. across and 4 in. deep. Put the vertex at the origin, so a rim point is $(8,4)$ . Use $x^2=4py$ :

$8^2=4p(4)\;\Rightarrow\;64=16p\;\Rightarrow\;p=4$

The food belongs 4 in. above the vertex (p. 631). The same method for a 100 m wide, 20 m deep telescope uses the point $(50,20)$ . It gives $2500=80p$ , so $p=31.25$ m (p. 631).

Solving Non-Linear Systems

To find where two curves cross, solve their equations together. Substitution and elimination both still work (p. 632).

Substitution. Find the intersection of $\frac{x^2}{4}+\frac{y^2}{25}=1$ and $x^2+y^2=9$ .

Clear fractions in the ellipse by multiplying by 100: $25x^2+4y^2=100$ .
Solve the circle for $x^2$ (not for $x$ ): $x^2=9-y^2$ .
Substitute: $25(9-y^2)+4y^2=100$ , so $-21y^2=-125$ and $y^2=\frac{125}{21}$ .
Back-substitute: $x^2=9-\frac{125}{21}=\frac{64}{21}$ .

Each square root has a plus and a minus, so $x=\pm\frac{8}{\sqrt{21}}$ and $y=\pm\frac{5\sqrt5}{\sqrt{21}}$ . That gives four intersection points (p. 632).

Elimination. Multiply the circle by $-4$ to get $-4x^2-4y^2=-36$ . Add it to $25x^2+4y^2=100$ . The $y^2$ terms cancel, leaving $21x^2=64$ , the same $x$ -values as before (p. 633).

A quadratic in disguise. Substitute $y=2x^2$ into $\frac{y^2}{4}-\frac{x^2}{9}=1$ . This gives $x^4-\frac{x^2}{9}=1$ , or

$9x^4-x^2-9=0$

Only even powers appear, so let $u=x^2$ . Then $9u^2-u-9=0$ and the quadratic formula gives $u=\frac{1\pm\sqrt{325}}{18}$ . Since $u=x^2$ cannot be negative, discard the minus root. Then $u\approx1.057$ , so $x\approx\pm1.028$ . Then $y=2x^2\approx2.114$ . The curves meet at about $(1.028,\,2.114)$ and $(-1.028,\,2.114)$ (p. 633).

Locating a Ship with Two Hyperbolas

Each pair of LORAN stations gives one hyperbola, and the ship lies on both. In the example, the first pair gives $\frac{x^2}{2025}-\frac{y^2}{3600}=1$ . A second pair 200 km farther south gives $\frac{x^2}{225}-\frac{(y+200)^2}{2275}=1$ (pp. 634–635).

Solve each equation for $x^2$ :

$x^2=2025+\frac{9}{16}y^2\qquad x^2=225+\frac{9}{91}(y+200)^2$

Set them equal and simplify. This gives the quadratic $75y^2-6400y-348800=0$ , with roots $y\approx123.11$ km and $y\approx-37.78$ km (p. 635).

Putting each $y$ back into a hyperbola gives $x\approx\pm102.71$ for the first root and $x\approx\pm53.18$ for the second. That makes four candidate positions. Two lie on land, so they are discarded. Navigators use other methods to choose between the remaining two (p. 635). In general, intersecting curves can give several solutions, and the real-world setting decides which one is correct.

One definition for all conics

Earlier sections defined each conic in its own way. A single rule covers ellipses, parabolas and hyperbolas. Put a focus $F$ at the origin and draw a line $L$ , the directrix, at $x=\pm p$ or $y=\pm p$ . A conic is the set of points $Q$ where the ratio of the distance to the focus over the distance to the directrix is a constant $e$ :

$\frac{d(Q,F)}{d(Q,L)}=e,\quad e>0$

The constant $e$ is the eccentricity. (p. 640)

Watch out: this $e$ is not the number $2.71828\ldots$ from exponential functions. Here it only names a distance ratio, and it can be any positive number, such as $0.5$ or $3$ . (p. 640)

The polar equation of a conic

Write $Q$ in polar form as $(r\cos\theta, r\sin\theta)$ . Take the directrix $y=-p$ . The distance from $Q$ to the focus is just $r$ . The distance from $Q$ to the line is $p+r\sin\theta$ . So the rule becomes $\frac{r}{p+r\sin\theta}=e$ . Solving for $r$ :

Clear the fraction: $r=e(p+r\sin\theta)$ .
Gather the $r$ terms: $r-er\sin\theta=ep$ .
Factor: $r(1-e\sin\theta)=ep$ , so $r=\frac{ep}{1-e\sin\theta}$ . (pp. 640–641)

The other three directrix positions work the same way:

Horizontal directrix $y=\pm p$ : $r=\frac{ep}{1\pm e\sin\theta}$ .
Vertical directrix $x=\pm p$ : $r=\frac{ep}{1\pm e\cos\theta}$ .

The sign matches the side of the origin the directrix is on. A plus sign means the directrix is at a positive value ( $y=p$ or $x=p$ ). A minus sign means it is at a negative value ( $y=-p$ or $x=-p$ ). (pp. 640–641, 644)

Writing a polar equation from $e$ and the directrix

Pick sine for a horizontal directrix or cosine for a vertical one.
Pick the sign from the side of the origin the directrix is on.
Compute the numerator $ep$ .

Example: $e=3$ and directrix $x=2$ . The directrix is vertical and at positive $x$ , so use cosine with a plus sign. Here $ep=3\cdot2=6$ , so

$r=\frac{6}{1+3\cos\theta}$

This graphs as a hyperbola. (p. 641)

How eccentricity sets the shape

$0<e<1$ : ellipse
$e=1$ : parabola
$e>1$ : hyperbola (p. 642)

Why it works (algebra). Take the directrix $x=1$ , so $r=\frac{e}{1+e\cos\theta}$ . Multiply out and use $r\cos\theta=x$ and $r^2=x^2+y^2$ . This gives

$(1-e^2)x^2+y^2+2e^2x=e^2$

If $e<1$ , the $x^2$ and $y^2$ coefficients are both positive, which is an ellipse. If $e>1$ , the $x^2$ coefficient is negative, which is a hyperbola. If $e=1$ , the $x^2$ term vanishes, which is a parabola. (p. 642)

Why it works (intuition). Look at the denominator $1+e\cos\theta$ .

If $e<1$ , it is always positive, so $r$ is finite at every angle. That closed loop is an ellipse.
If $e=1$ , it hits zero at one angle ( $\theta=\pi$ ), so $r$ is undefined there. That matches a parabola, which has no point in the direction of its axis away from the vertex.
If $e>1$ , it is zero at two angles and $r$ turns negative for some angles. The positive and negative parts make two separate branches, which is a hyperbola. (p. 643)

Identifying a conic from its polar equation

Divide the top and bottom so the denominator starts with $1$ .
Read $e$ as the coefficient of $\sin\theta$ or $\cos\theta$ .
Use the numerator $ep$ to find $p$ .
Use the function and sign to name the directrix.

Example A: $r=\frac{6}{3-2\cos\theta}$ . Divide by 3 to get $\frac{2}{1-\frac23\cos\theta}$ . Then $e=\frac23$ , so it is an ellipse. From $ep=2$ , $p=3$ . The minus sign gives directrix $x=-3$ . (pp. 643–644)

Example B: $r=\frac{8}{1-2\sin\theta}$ is already in standard form. Here $e=2$ , so it is a hyperbola. From $ep=8$ , $p=4$ , and the directrix is $y=-4$ . (p. 643)

Example C: $r=\frac{8}{5+5\sin\theta}$ . Divide by 5 to get $\frac{8/5}{1+\sin\theta}$ . Then $e=1$ , so it is a parabola, and $p=\frac85$ . The directrix $y=\frac85$ lies above the focus, so the parabola opens downward, away from the directrix. (p. 644)

Sketching a polar ellipse

Evaluate $r$ at $\theta=0,\frac{\pi}{2},\pi,\frac{3\pi}{2}$ and plot the points. Then use the vertices to find the center, $a$ , $c$ and $b$ . Remember that the origin is one focus.

Example: $r=\frac{3}{1-0.5\sin\theta}$ . Here $e=0.5$ , an ellipse, and $ep=3$ gives $p=6$ , so the directrix is $y=-6$ .

The points are $(3,0)$ , $(0,6)$ , $(-3,0)$ and $(0,-2)$ .
The vertices are $(0,-2)$ and $(0,6)$ , so the center is the midpoint $(0,2)$ and $a=4$ .
The focus at the origin is $2$ from the center, so $c=2$ .
Then $b^2=a^2-c^2=16-4=12$ .
The Cartesian equation is $\frac{x^2}{12}+\frac{(y-2)^2}{16}=1$ . (pp. 645–646)

Analyzing a polar hyperbola

Find the vertices on the axis. Their midpoint is the center. The distance from the center to the origin focus is $c$ , and $b^2=c^2-a^2$ .

Example: $r=\frac{6}{1+2\cos\theta}$ . Here $e=2$ , so it is a hyperbola.

At $\theta=0$ , $r=2$ , giving $(2,0)$ . At $\theta=\pi$ , $r=-6$ , which plots at $(6,0)$ .
The vertices are $(2,0)$ and $(6,0)$ , so the center is $(4,0)$ and $a=2$ .
The origin focus is $4$ from the center, so $c=4$ and $b^2=16-4=12$ .
The equation is $\frac{(x-4)^2}{4}-\frac{y^2}{12}=1$ .
The asymptotes are $y=\pm\sqrt3\,(x-4)$ . (pp. 646–647)

Check yourself

For an ellipse with $a=5$ and $b=3$ , find $c$ . Why is $b^2=a^2-c^2$ not simply the Pythagorean Theorem applied to any right triangle?
A hyperbola opens horizontally with center $(2,-1)$ , $a=3$ and $b=4$ . Write its equation and the equations of its asymptotes.
A parabolic dish has vertex at the origin and passes through the rim point $(4,2)$ on $x^2=4py$ . How do you find where to place the receiver (the focus)?
Write the polar equation of the conic with eccentricity $\frac12$ and directrix $y=4$ . What shape is it?
A system has the equations $x^2+y^2=25$ and $y=x^2-5$ . Describe how you would solve it by substitution.

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