Conic Sections

Conic sections have been studied since the time of the ancient Greeks, and were considered to be an important mathematical concept. As early as 320 BCE, such Greek mathematicians as Menaechmus, Appollonius, and Archimedes were fascinated by these curves. Appollonius wrote an entire eight-volume treatise on conic sections in which he was, for example, able to derive a specific method for identifying a conic section through the use of geometry. Since then, important applications of conic sections have arisen (for example, in astronomy), and the properties of conic sections are used in radio telescopes, satellite dish receivers, and even architecture. In this section we discuss the three basic conic sections, some of their properties, and their equations.

Conic sections get their name because they can be generated by intersecting a plane with a cone. A cone has two identically shaped parts called nappes. One nappe is what most people mean by “cone,” having the shape of a party hat. A right circular cone can be generated by revolving a line passing through the origin around the y-axis as shown.

Conic sections are generated by the intersection of a plane with a cone ([link]). If the plane is parallel to the axis of revolution (the y-axis), then the conic section is a hyperbola. If the plane is parallel to the generating line, the conic section is a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects one nappe at an angle to the axis (other than

90 °),

Parabolas

A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects only one of the nappes. A parabola can also be defined in terms of distances.

A graph of a typical parabola appears in [link]. Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point P with coordinates

(x_{1}, y_{1})

to denote the coordinates of the vertex. Then if the focus is directly above the vertex, it has coordinates

(h, k + p)

Going through the same derivation yields the formula

{(x - h)}^{2} = 4 p (y - k) .

We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.

In addition, the equation of a parabola can be written in the general form, though in this form the values of h, k, and p are not immediately recognizable. The general form of a parabola is written as

The first equation represents a parabola that opens either up or down. The second equation represents a parabola that opens either to the left or to the right. To put the equation into standard form, use the method of completing the square.

Converting the Equation of a Parabola from General into Standard Form

Put the equation $x^{2} - 4 x - 8 y + 12 = 0$

into standard form and graph the resulting parabola.

Since y is not squared in this equation, we know that the parabola opens either upward or downward. Therefore we need to solve this equation for y, which will put the equation into standard form. To do that, first add $8 y$

to both sides of the equation:

8 y = x^{2} - 4 x + 12 .

The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the right-hand side using parentheses:

8 y = (x^{2} - 4 x) + 12 .

Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses a perfect square trinomial. To do this, take half the coefficient of x and square it. This gives ${(\frac{−4}{2})}^{2} = 4 .$

Add 4 inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:

8 y = (x^{2} - 4 x + 4) + 12 - 4 .

Now combine like terms and factor the quantity inside the parentheses:

8 y = {(x - 2)}^{2} + 8 .

Finally, divide by 8:

y = \frac{1}{8} {(x - 2)}^{2} + 1 .

This equation is now in standard form. Comparing this to [link] gives $h = 2,$

k = 1,

and $p = 2 .$

The parabola opens up, with vertex at $(2, 1),$

focus at $(2, 3),$

and directrix $y = −1 .$

The graph of this parabola appears as follows.

A parabola is drawn with vertex at (2, 1) and opening up with equation x2 – 4x – 8y + 12 = 0. The focus is drawn at (1, 3). The directrix is drawn at y = − 1.

The axis of symmetry of a vertical (opening up or down) parabola is a vertical line passing through the vertex. The parabola has an interesting reflective property. Suppose we have a satellite dish with a parabolic cross section. If a beam of electromagnetic waves, such as light or radio waves, comes into the dish in a straight line from a satellite (parallel to the axis of symmetry), then the waves reflect off the dish and collect at the focus of the parabola as shown.

A parabola is drawn with vertex at the origin and opening up. Two parallel lines are drawn that strike the parabola and reflect to the focus.

Consider a parabolic dish designed to collect signals from a satellite in space. The dish is aimed directly at the satellite, and a receiver is located at the focus of the parabola. Radio waves coming in from the satellite are reflected off the surface of the parabola to the receiver, which collects and decodes the digital signals. This allows a small receiver to gather signals from a wide angle of sky. Flashlights and headlights in a car work on the same principle, but in reverse: the source of the light (that is, the light bulb) is located at the focus and the reflecting surface on the parabolic mirror focuses the beam straight ahead. This allows a small light bulb to illuminate a wide angle of space in front of the flashlight or car.

Ellipses

An ellipse can also be defined in terms of distances. In the case of an ellipse, there are two foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more detail later in this section.

A graph of a typical ellipse is shown in [link]. In this figure the foci are labeled as

F

Both are the same fixed distance from the origin, and this distance is represented by the variable c. Therefore the coordinates of

F

are located at the ends of the major axis of the ellipse, and have coordinates

(a, 0)

respectively. The major axis is always the longest distance across the ellipse, and can be horizontal or vertical. Thus, the length of the major axis in this ellipse is 2a. Furthermore,

P

are located at the ends of the minor axis of the ellipse, and have coordinates

(0, b)

respectively. The minor axis is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis.

According to the definition of the ellipse, we can choose any point on the ellipse and the sum of the distances from this point to the two foci is constant. Suppose we choose the point P. Since the coordinates of point P are

(a, 0),

Therefore the sum of the distances from an arbitrary point A with coordinates

(x, y)

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

If we refer back to [link], then the length of each of the two green line segments is equal to a. This is true because the sum of the distances from the point Q to the foci

F and F^{'}

is equal to 2a, and the lengths of these two line segments are equal. This line segment forms a right triangle with hypotenuse length a and leg lengths b and c. From the Pythagorean theorem,

a^{2} + b^{2} = c^{2}

Finally, if the center of the ellipse is moved from the origin to a point

(h, k),

If the major axis is horizontal, then the ellipse is called horizontal, and if the major axis is vertical, then the ellipse is called vertical. The equation of an ellipse is in general form if it is in the form

A x^{2} + B y^{2} + C x + D y + E = 0,

where A and B are either both positive or both negative. To convert the equation from general to standard form, use the method of completing the square.

Finding the Standard Form of an Ellipse

Put the equation $9 x^{2} + 4 y^{2} - 36 x + 24 y + 36 = 0$

into standard form and graph the resulting ellipse.

First subtract 36 from both sides of the equation:

9 x^{2} + 4 y^{2} - 36 x + 24 y = −36 .

Next group the x terms together and the y terms together, and factor out the common factor:

\begin{matrix} (9 x^{2} - 36 x) + (4 y^{2} + 24 y) & = & −36 \\ 9 (x^{2} - 4 x) + 4 (y^{2} + 6 y) & = & −36. \end{matrix}

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives ${(\frac{−4}{2})}^{2} = 4 .$

In the second set of parentheses, take half the coefficient of y and square it. This gives ${(\frac{6}{2})}^{2} = 9 .$

Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are adding 36 to the second set as well. Therefore the equation becomes

\begin{matrix} 9 (x^{2} - 4 x + 4) + 4 (y^{2} + 6 y + 9) = −36 + 36 + 36 \\ 9 (x^{2} - 4 x + 4) + 4 (y^{2} + 6 y + 9) = 36. \end{matrix}

Now factor both sets of parentheses and divide by 36:

\begin{matrix} 9 {(x - 2)}^{2} + 4 {(y + 3)}^{2} & = & 36 \\ \frac{9 {(x - 2)}^{2}}{36} + \frac{4 {(y + 3)}^{2}}{36} & = & 1 \\ \frac{{(x - 2)}^{2}}{4} + \frac{{(y + 3)}^{2}}{9} & = & 1. \end{matrix}

The equation is now in standard form. Comparing this to [link] gives $h = 2,$

k = −3,

a = 3,

and $b = 2 .$

This is a vertical ellipse with center at $(2, −3),$

major axis 6, and minor axis 4. The graph of this ellipse appears as follows.

An ellipse is drawn with equation 9x2 + 4y2 – 36x + 24y + 36 = 0. It has center at (2, −3), touches the x axis at (2, 0), and touches the y axis at (0, −3).

According to Kepler’s first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one of the foci as shown in [link](a). Because Earth’s orbit is an ellipse, the distance from the Sun varies throughout the year. A commonly held misconception is that Earth is closer to the Sun in the summer. In fact, in summer for the northern hemisphere, Earth is farther from the Sun than during winter. The difference in season is caused by the tilt of Earth’s axis in the orbital plane. Comets that orbit the Sun, such as Halley’s Comet, also have elliptical orbits, as do moons orbiting the planets and satellites orbiting Earth.

Ellipses also have interesting reflective properties: A light ray emanating from one focus passes through the other focus after mirror reflection in the ellipse. The same thing occurs with a sound wave as well. The National Statuary Hall in the U.S. Capitol in Washington, DC, is a famous room in an elliptical shape as shown in [link](b). This hall served as the meeting place for the U.S. House of Representatives for almost fifty years. The location of the two foci of this semi-elliptical room are clearly identified by marks on the floor, and even if the room is full of visitors, when two people stand on these spots and speak to each other, they can hear each other much more clearly than they can hear someone standing close by. Legend has it that John Quincy Adams had his desk located on one of the foci and was able to eavesdrop on everyone else in the House without ever needing to stand. Although this makes a good story, it is unlikely to be true, because the original ceiling produced so many echoes that the entire room had to be hung with carpets to dampen the noise. The ceiling was rebuilt in 1902 and only then did the now-famous whispering effect emerge. Another famous whispering gallery—the site of many marriage proposals—is in Grand Central Station in New York City.

Hyperbolas

A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.

The derivation of the equation of a hyperbola in standard form is virtually identical to that of an ellipse. One slight hitch lies in the definition: The difference between two numbers is always positive. Let P be a point on the hyperbola with coordinates

(x, y) .

Then the definition of the hyperbola gives

\| d (P, F_{1}) - d (P, F_{2}) \| = constant .

To simplify the derivation, assume that P is on the right branch of the hyperbola, so the absolute value bars drop. If it is on the left branch, then the subtraction is reversed. The vertex of the right branch has coordinates

(a, 0),

This equation is therefore true for any point on the hyperbola. Returning to the coordinates

(x, y)

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

Finally, if the center of the hyperbola is moved from the origin to the point

(h, k),

If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general form if it is in the form

A x^{2} + B y^{2} + C x + D y + E = 0,

where A and B have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.

Finding the Standard Form of a Hyperbola

Put the equation $9 x^{2} - 16 y^{2} + 36 x + 32 y - 124 = 0$

into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

First add 124 to both sides of the equation:

9 x^{2} - 16 y^{2} + 36 x + 32 y = 124 .

Next group the x terms together and the y terms together, then factor out the common factors:

\begin{matrix} (9 x^{2} + 36 x) - (16 y^{2} - 32 y) & = & 124 \\ 9 (x^{2} + 4 x) - 16 (y^{2} - 2 y) & = & 124. \end{matrix}

In the second set of parentheses, take half the coefficient of y and square it. This gives ${(\frac{−2}{2})}^{2} = 1 .$

Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes

\begin{matrix} 9 (x^{2} + 4 x + 4) - 16 (y^{2} - 2 y + 1) = 124 + 36 - 16 \\ 9 (x^{2} + 4 x + 4) - 16 (y^{2} - 2 y + 1) = 144. \end{matrix}

Next factor both sets of parentheses and divide by 144:

\begin{matrix} 9 {(x + 2)}^{2} - 16 {(y - 1)}^{2} & = & 144 \\ \frac{9 {(x + 2)}^{2}}{144} - \frac{16 {(y - 1)}^{2}}{144} & = & 1 \\ \frac{{(x + 2)}^{2}}{16} - \frac{{(y - 1)}^{2}}{9} & = & 1. \end{matrix}

The equation is now in standard form. Comparing this to [link] gives $h = −2,$

k = 1,

a = 4,

and $b = 3 .$

This is a horizontal hyperbola with center at $(−2, 1)$

and asymptotes given by the equations $y = 1 \pm \frac{3}{4} (x + 2) .$

The graph of this hyperbola appears in the following figure.

A hyperbola is drawn with equation 9x2 + 16y2 + 36x + 32y – 124 = 0. It has center at (−2, 1), and the hyperbolas are open to the left and right.

Hyperbolas also have interesting reflective properties. A ray directed toward one focus of a hyperbola is reflected by a hyperbolic mirror toward the other focus. This concept is illustrated in the following figure.

This property of the hyperbola has important applications. It is used in radio direction finding (since the difference in signals from two towers is constant along hyperbolas), and in the construction of mirrors inside telescopes (to reflect light coming from the parabolic mirror to the eyepiece). Another interesting fact about hyperbolas is that for a comet entering the solar system, if the speed is great enough to escape the Sun’s gravitational pull, then the path that the comet takes as it passes through the solar system is hyperbolic.

Eccentricity and Directrix

An alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. We will see that the value of the eccentricity of a conic section can uniquely define that conic.

Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the directrices of a horizontal ellipse are

x = ± \frac{a^{2}}{c} .

and the distance from the vertex to the right directrix is

\frac{a^{2}}{c} - c .

this step proves that the eccentricity of an ellipse is less than 1. The directrices of a horizontal hyperbola are also located at

x = ± \frac{a^{2}}{c},

and a similar calculation shows that the eccentricity of a hyperbola is also

e = \frac{c}{a} .

Polar Equations of Conic Sections

Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section p is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where a is the length of the semi-major axis (i.e., half the length of the major axis), c is the distance from the origin to the focus, and e is the eccentricity. In the case of a parabola, a represents the distance from the vertex to the focus.

Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.

Eccentricities and Focal Parameters of the Conic Sections
Conic	e	p
Ellipse	$0 < e < 1$	$\frac{a^{2} - c^{2}}{c} = \frac{a (1 - e^{2})}{c}$
Parabola	$e = 1$	$2 a$
Hyperbola	$e > 1$	$\frac{c^{2} - a^{2}}{c} = \frac{a (e^{2} - 1)}{e}$

In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar form, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is at the origin only if the conic is a circle (i.e.,

e = 0) .

Graphing a Conic Section in Polar Coordinates

Identify and create a graph of the conic section described by the equation

r = \frac{3}{1 + 2 cos θ} .

The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter p can be calculated by using the equation $e p = 3 .$

Since $e = 2,$

this gives $p = \frac{3}{2} .$

The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for $θ$

and create a table of values. Then we can graph the hyperbola ([link]).

θ

r

θ

r


{: valign=”top”}	———-
0	1	$π$

−3
{: valign=”top”}	$\frac{π}{4}$

\frac{3}{1 + \sqrt{2}} \approx 1.2426

\frac{5 π}{4}

\frac{3}{1 - \sqrt{2}} \approx −7.2426


{: valign=”top”}	$\frac{π}{2}$

3	$\frac{3 π}{2}$

3
{: valign=”top”}	$\frac{3 π}{4}$

\frac{3}{1 - \sqrt{2}} \approx −7.2426

\frac{7 π}{4}

\frac{3}{1 + \sqrt{2}} \approx 1.2426

| {: valign=”top”}{: .unnumbered summary=”This table has two columns and nine rows. The first row is a header row and reads from left to right θ and 4. After the header, the first column reads 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, and 7π/4. The second column reads 1, 3 divided by the quantity (1 + the square root of 2), which is approximately equal to 1.2426, 3, 3 divided by the quantity (1 – the square root of 2), which is approximately equal to −7.2426, −3, 3 divided by the quantity (1 – the square root of 2), which is approximately equal to −7.2426, 3, and 3 divided by the quantity (1 + the square root of 2), which is approximately equal to 1.2426.” data-label=””}

Graph of a hyperbola with equation r = 3/(1 + 2 cosθ), center at (2, 0), and vertices at (1, 0) and (3, 0).

General Equations of Degree Two

then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section

4 A C - B^{2} .

The simplest example of a second-degree equation involving a cross term is

x y = 1 .

The asymptotes of this hyperbola are the x and y coordinate axes. To determine the angle

θ

of rotation of the conic section, we use the formula

cot 2 θ = \frac{A - C}{B} .

The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled

A^{'}, B^{'}, C^{'}, D^{'}, E^{'}, and F^{'},

Identifying a Rotated Conic

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

13 x^{2} - 6 \sqrt{3} x y + 7 y^{2} - 256 = 0 .

In this equation, $A = 13, B = −6 \sqrt{3}, C = 7, D = 0, E = 0,$

and $F = −256 .$

The discriminant of this equation is $4 A C - B^{2} = 4 (13) (7) - {(−6 \sqrt{3})}^{2} = 364 - 108 = 256 .$

Therefore this conic is an ellipse. To calculate the angle of rotation of the axes, use $cot 2 θ = \frac{A - C}{B} .$

This gives

\begin{matrix} cot 2 θ & = \frac{A - C}{B} \\ = \frac{13 - 7}{−6 \sqrt{3}} \\ = - \frac{\sqrt{3}}{3} . \end{matrix}

Therefore $2 θ = 120^{o}$

and $θ = 60^{o},$

which is the angle of the rotation of the axes.

To determine the rotated coefficients, use the formulas given above:

\begin{matrix} A^{'} & = & A {cos}^{2} θ + B cos θ sin θ + C {sin}^{2} θ \\ = & 13 {cos}^{2} 60 + (−6 \sqrt{3}) cos 60 sin 60 + 7 {sin}^{2} 60 \\ = & 13 {(\frac{1}{2})}^{2} - 6 \sqrt{3} (\frac{1}{2}) (\frac{\sqrt{3}}{2}) + 7 {(\frac{\sqrt{3}}{2})}^{2} \\ = & 4, \\ B^{'} & = & 0, \\ C^{'} & = & A {sin}^{2} θ - B sin θ cos θ + C {cos}^{2} θ \\ = & 13 {sin}^{2} 60 + (−6 \sqrt{3}) sin 60 cos 60 = 7 {cos}^{2} 60 \\ = & {(\frac{\sqrt{3}}{2})}^{2} + 6 \sqrt{3} (\frac{\sqrt{3}}{2}) (\frac{1}{2}) + 7 {(\frac{1}{2})}^{2} \\ = & 16, \\ D^{'} & = & D cos θ + E sin θ \\ = & (0) cos 60 + (0) sin 60 \\ = & 0, \\ E^{'} & = & − D sin θ + E cos θ \\ = & − (0) sin 60 + (0) cos 60 \\ = & 0, \\ F^{'} & = & F \\ = & −256. \end{matrix}

The equation of the conic in the rotated coordinate system becomes

\begin{matrix} 4 {(x^{'})}^{2} + 16 {(y^{'})}^{2} & = & 256 \\ \frac{{(x^{'})}^{2}}{64} + \frac{{(y^{'})}^{2}}{16} & = & 1. \end{matrix}

A graph of this conic section appears as follows.

Graph of an ellipse with equation 13x2 – 6 times the square root of 3 times xy + 7y2 – 256 = 0. The center is at the origin, and the ellipse appears to be skewed 60 degrees. There are dashed red lines along the major and minor axes.

Key Concepts

For the following exercises, determine the equation of the parabola using the information given.

Focus $(4, 0)$

and directrix $x = −4$

y^{2} = 16 x

Focus $(0, −3)$

and directrix $y = 3$

Focus $(0, 0.5)$

and directrix $y = −0.5$

x^{2} = 2 y

Focus $(2, 3)$

and directrix $x = −2$

Focus $(0, 2)$

and directrix $y = 4$

x^{2} = −4 (y - 3)

Focus $(−1, 4)$

and directrix $x = 5$

Focus $(−3, 5)$

and directrix $y = 1$

{(x + 3)}^{2} = 8 (y - 3)

Focus $(\frac{5}{2}, −4)$

and directrix $x = \frac{7}{2}$

For the following exercises, determine the equation of the ellipse using the information given.

Endpoints of major axis at $(4, 0), (−4, 0)$

and foci located at $(2, 0), (−2, 0)$

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

Endpoints of major axis at $(0, 5), (0, −5)$

and foci located at $(0, 3), (0, −3)$

Endpoints of major axis at $(0, 2), (0, −2)$

and foci located at $(3, 0), (−3, 0)$

\frac{x^{2}}{13} + \frac{y^{2}}{4} = 1

Endpoints of major axis at $(−3, 3), (7, 3)$

and foci located at $(−2, 3), (6, 3)$

Endpoints of major axis at $(−3, 5), (−3, −3)$

and foci located at $(−3, 3), (−3, −1)$

\frac{{(y - 1)}^{2}}{16} + \frac{{(x + 3)}^{2}}{12} = 1

Endpoints of major axis at $(0, 0), (0, 4)$

and foci located at $(5, 2), (−5, 2)$

Foci located at $(2, 0), (−2, 0)$

and eccentricity of $\frac{1}{2}$

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

Foci located at $(0, −3), (0, 3)$

and eccentricity of $\frac{3}{4}$

For the following exercises, determine the equation of the hyperbola using the information given.

Vertices located at $(5, 0), (−5, 0)$

and foci located at $(6, 0), (−6, 0)$

\frac{x^{2}}{25} - \frac{y^{2}}{11} = 1

Vertices located at $(0, 2), (0, −2)$

and foci located at $(0, 3), (0, −3)$

Endpoints of the conjugate axis located at $(0, 3), (0, −3)$

and foci located $(4, 0), (−4, 0)$

\frac{x^{2}}{7} - \frac{y^{2}}{9} = 1

Vertices located at $(0, 1), (6, 1)$

and focus located at $(8, 1)$

Vertices located at $(−2, 0), (−2, −4)$

and focus located at $(−2, −8)$

\frac{{(y + 2)}^{2}}{4} - \frac{{(x + 2)}^{2}}{32} = 1

Endpoints of the conjugate axis located at $(3, 2), (3, 4)$

and focus located at $(3, 7)$

Foci located at $(6, −0), (6, 0)$

and eccentricity of 3

\frac{x^{2}}{4} - \frac{y^{2}}{32} = 1

(0, 10), (0, −10)

and eccentricity of 2.5

For the following exercises, consider the following polar equations of conics. Determine the eccentricity and identify the conic.

r = \frac{−1}{1 + cos θ}

e = 1,

parabola

r = \frac{8}{2 - sin θ}

r = \frac{5}{2 + sin θ}

e = \frac{1}{2},

ellipse

r = \frac{5}{−1 + 2 sin θ}

r = \frac{3}{2 - 6 sin θ}

e = 3,

hyperbola

r = \frac{3}{−4 + 3 sin θ}

For the following exercises, find a polar equation of the conic with focus at the origin and eccentricity and directrix as given.

Directrix: x = 4; e = \frac{1}{5}

r = \frac{4}{5 + cos θ}

Directrix: x = −4; e = 5

Directrix: y = 2; e = 2

r = \frac{4}{1 + 2 sin θ}

Directrix: y = −2; e = \frac{1}{2}

For the following exercises, sketch the graph of each conic.

r = \frac{1}{1 + sin θ}

Graph of a parabola open down with center at the origin.

r = \frac{1}{1 - cos θ}

r = \frac{4}{1 + cos θ}

Graph of a parabola open to the left with center near the origin.

r = \frac{10}{5 + 4 sin θ}

r = \frac{15}{3 - 2 cos θ}

Graph of an ellipse with center near (8, 0), major axis horizontal and roughly 18, and minor axis slightly more than 12.

r = \frac{32}{3 + 5 sin θ}

r (2 + sin θ) = 4

Graph of an circle with center near (0, −1.5) and radius near 2.5.

r = \frac{3}{2 + 6 sin θ}

r = \frac{3}{−4 + 2 sin θ}

Graph of a circle with center (0, −0.5) and radius 1.

\begin{matrix} \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \end{matrix}

\frac{x^{2}}{4} + \frac{y^{2}}{16} = 1

Graph of an ellipse with center the origin and with major axis vertical and of length 8 and minor axis of length 4.

4 x^{2} + 9 y^{2} = 36

25 x^{2} - 4 y^{2} = 100

Graph of a hyperbola with center the origin and with the two halves open to the left and right. The vertices are on the x axis at ±2.

\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1

x^{2} = 12 y

Graph of a parabola with vertex the origin and open up.

y^{2} = 20 x

12 x = 5 y^{2}

Graph of a parabola with vertex the origin and open to the right.

For the following equations, determine which of the conic sections is described.

x y = 4

x^{2} + 4 x y - 2 y^{2} - 6 = 0

Hyperbola

x^{2} + 2 \sqrt{3} x y + 3 y^{2} - 6 = 0

x^{2} - x y + y^{2} - 2 = 0

Ellipse

34 x^{2} - 24 x y + 41 y^{2} - 25 = 0

52 x^{2} - 72 x y + 73 y^{2} + 40 x + 30 y - 75 = 0

Ellipse

The mirror in an automobile headlight has a parabolic cross section, with the lightbulb at the focus. On a schematic, the equation of the parabola is given as $x^{2} = 4 y .$

At what coordinates should you place the lightbulb?

A satellite dish is shaped like a paraboloid of revolution. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?

At the point 2.25 feet above the vertex.

Consider the satellite dish of the preceding problem. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver?

A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth.

0.5625 feet

Whispering galleries are rooms designed with elliptical ceilings. A person standing at one focus can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet and the foci are located 30 feet from the center, find the height of the ceiling at the center.

A person is standing 8 feet from the nearest wall in a whispering gallery. If that person is at one focus and the other focus is 80 feet away, what is the length and the height at the center of the gallery?

Length is 96 feet and height is approximately 26.53 feet.

For the following exercises, determine the polar equation form of the orbit given the length of the major axis and eccentricity for the orbits of the comets or planets. Distance is given in astronomical units (AU).

Halley’s Comet: length of major axis = 35.88, eccentricity = 0.967

Hale-Bopp Comet: length of major axis = 525.91, eccentricity = 0.995

r = \frac{2.616}{1 + 0.995 cos θ}

Mars: length of major axis = 3.049, eccentricity = 0.0934

Jupiter: length of major axis = 10.408, eccentricity = 0.0484

r = \frac{5.192}{1 + 0.0484 cos θ}

Chapter Review Exercises

For the following exercises, sketch the parametric curve and eliminate the parameter to find the Cartesian equation of the curve.

For the following exercises, sketch the polar curve and determine what type of symmetry exists, if any.

For the following exercises, find the polar equation for the curve given as a Cartesian equation.

For the following exercises, find the equation of the tangent line to the given curve. Graph both the function and its tangent line.

For the following exercises, find the arc length of the curve over the given interval.

For the following exercises, find the Cartesian equation describing the given shapes.

For the following exercises, determine the eccentricity and identify the conic. Sketch the conic.