Pre-Calculus 12 - Weebly · Pre-Calculus 12 Unit 8 –Conic Sections. 1. Introduction to conic...

Pre-Calculus 12Unit 8 – Conic Sections

1. Introduction to conic sections• A cone is a geometric figure that is created by the following steps:

1. Let 𝑎 and 𝑔 be the axis and generator, respectively, that intersect at a point 𝑉, called vertex.

2. Keep 𝑎 fixed, and rotate 𝑔 about 𝑎 while maintaining the same angle between 𝑎and 𝑔.

3. The collection of points swept out by 𝑔 is called a right circular cone.

4. A cone consists of two parts, called nappes, that intersect at 𝑉.

Unit 8 - Conic Sections 2

• When a plane intersects with a right circular cone, curves are formed. These curves are called conic sections.

• Depending on the orientation of the plane relative to the axis of the cone, four types of conic sections may result.


• Types of the intersection curve between a cone and a plane:


Type of the curve

Orientation of the plane Applications

Circle Angle with axis = 90° Ferris wheels, trigonometry

Ellipse Angle with axis > Angle between axis and generator

Orbits of planets, whisper chamber

Parabola Angle with axis = Angle between axis and generator

Satellite dishes, telescopes

Hyperbola Angle with axis < Angle between axis and generator

Nuclear cooling tower, navigation of ships and planes

• Conic sections can be described mathematically by the following equations called the general form:

• The type of a conic section is determined by the discriminant:


𝐴𝑥2 + 𝐵𝑥𝑦 + 𝐶𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0

Discriminant Type of conics

𝐵2 − 4𝐴𝐶 < 0 Ellipse

𝐵2 − 4𝐴𝐶 < 0 and 𝐵 = 0, 𝐴 = 𝐶 Circle

𝐵2 − 4𝐴𝐶 = 0 Parabola

𝐵2 − 4𝐴𝐶 > 0 Hyperbola

2. Circles• We have studied circles previously in the chapter of analytic geometry.

• Circle is defined as follows:

• Graphically:


• The simplest circle is the unit circle centered at the origin.

• The circle centered at the origin with a radius 𝑟 is given by:


𝑥2 + 𝑦2 = 𝑟2

The equation of the unit circle:

𝑥2 + 𝑦2 = 1

• For a generic circle of radius 𝑟 centered at (ℎ, 𝑘), its equation is:

This is called the standard form of a circle.


𝑥 − ℎ 2 + 𝑦 − 𝑘 2 = 𝑟2

• The standard form can be derived also by means of transformations – horizontal translation by ℎ and vertical translation by 𝑘.

• The general form and standard form of a circle can be interconverted.

• Example: The equation of a circle is (𝑥 + 3)2+ 𝑦 − 1 2 = 64.

a) Determine the radius and the coordinates of the center of the circle.

b) Sketch the circle and identify its domain and range.


• Example: A circle, centered at the origin, was translated. The equation of the image circle is:

a) Determine the radius of this circle, and the coordinates of its center.

b) Identify the translation.


𝑥2 + 𝑦2 + 12𝑥 − 6𝑦 + 20 = 0

• Example: Write an equation to describe each circle.

(a) (b)


• Example: A circle has a diameter 𝑃𝑄 with endpoints 𝑃(−5,−7) and 𝑄(3,−1). Write the equation of the circle in general form.


3. Ellipses

• When a circle is stretched or compressed in a particular direction, the image graph is an ellipse.

• Special features of ellipses:


• The geometric definition of an ellipse is:

• Schematically, for an ellipse centered at the origin:


𝑑1 + 𝑑2 = constant

• Let 2𝑎 be the constant; we have

• It can be shown that this equation is equivalent to the following standard form assuming that 𝑎2 − 𝑐2 = 𝑏2:

• Length of semi-major axis: 𝑎

• Length of semi-minor axis: 𝑏

• Locations of foci: ±𝑐, 0Unit 8 - Conic Sections 15

𝑥 + 𝑐 2 + 𝑦2 + 𝑥 − 𝑐 2 + 𝑦2 = 2𝑎

𝑥2

𝑎2+𝑦2

𝑏2= 1

• Two possible orientations:


Horizontal ellipse: 𝑎 > 𝑏 > 0 Vertical ellipse: 𝑏 > 𝑎 > 0

Focus: 𝑐 = ± 𝑎2 − 𝑏2 Focus: 𝑐 = ± 𝑏2 − 𝑎2


• Example: An ellipse has the equation

Find the foci, vertices, and the lengths of the major and minor axes.

𝑥2

9+𝑦2

4= 1

• Example: The vertices of an ellipse are (±4, 0), and the foci are (±2, 0). Find its equation, and sketch the graph.


• Example: Analyze the equation: 9𝑥2 + 𝑦2 = 9.


• When the graph of an ellipse is translated ℎ units right and 𝑘 units up, the equation of the image graph is:

• Graphically:


𝑥 − ℎ 2

𝑎2+

𝑦 − 𝑘 2

𝑏2= 1

• Example: An ellipse is described by the equation:

a) Write the equation is standard form. Identify the coordinates of the center of the ellipse.

b) Determine the directions and lengths of the major and minor axes.

c) Sketch the ellipse, then identify its domain and range.

d) Find the locations of foci.


16 𝑥 + 2 2 + 9 𝑦 − 5 2 = 144

• Example: An ellipse, centered at the origin, was translated. The equation of the image ellipse is:

a) Determine the coordinates of the center.

b) Identify the translation.


4𝑥2 + 9𝑦2 + 8𝑥 − 54𝑦 + 49 = 0

• We have seen that an ellipse results from stretching a circle in a particular direction. The degree of which an ellipse deviates from the parent circle is measured by eccentricity, which is defined as:


𝑒 =𝑐

𝑎=

𝑎2 − 𝑏2

𝑎= 1 −

𝑏2

𝑎2

• Example: Find the equation of the ellipse with foci (0, ±8) and eccentricity 𝑒 = 0.8, and sketch its graph.


• Example: Find an equation for the ellipse that has the eccentricity 𝑒 =3/2, foci on 𝑥-axis, and length of major axis 4.


4. Parabolas• We have seen in PREC11 the quadratic functions whose graphs are referred to

as parabolas:

• The vertex form:


𝑦 = 𝑎 𝑥 − ℎ 2 + 𝑘

• The vertex form can be converted to the standard form:

• For the “horizontal” parabolas:

• The vertex form:

• The standard form:


𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐

𝑥 = 𝑎 𝑦 − 𝑘 2 + ℎ

𝑥 = 𝑎𝑦2 + 𝑏𝑦 + 𝑐

• From a geometric point of view, a parabola can be defined as follows:

• Graphically:


• Consider a parabola with the vertex at the origin:

• By definition,

• It means that

• We can show that this expression is equivalent to:

which is a quadratic function!


|𝑀𝐹| = |𝑀𝐷|

𝑥2 + 𝑦 − 𝑝 2 = 𝑦 + 𝑝

𝑥2 = 4𝑝𝑦

• By the same token, we can show that for a parabola with the vertex at the origin and a horizontal axis of symmetry,

• The length of latus rectum is

• It is also called the focal diameter.


𝑦2 = 4𝑝𝑥

𝐿𝑅 = 4𝑝

• The sign of 𝑝 determines the orientation of a parabola.


• Example: Find the equation of the parabola with vertex (0, 0) and focus (0, 2). Sketch its graph and indicate the directrix.


• Example: A parabola has the equation 6𝑥 + 𝑦2 = 0. Find the focus and directrix of the parabola, and sketch the graph.


• Example: Find the equation of the parabola with directrix 𝑥 = 2 and focus (−2, 0). Graph the parabola.


• Example: Graph 𝑦 = 24𝑥2. Identify and label the focus, directrix, and endpoints of the latus rectum. Determine the focal diameter.


• Suppose that a parabola is shifted both horizontally and vertically so that its vertex is at the point ℎ, 𝑘 .


𝑥 − ℎ 2 = 4𝑝(𝑦 − 𝑘) 𝑦 − 𝑘 2 = 4𝑝(𝑥 − ℎ)

• Example: Determine the vertex, focus, and directrix and sketch the graph of the parabola.


𝑥2 − 4𝑥 = 8𝑦 − 28

• Example: Find the vertex, focus, axis, directrix, and graph of the parabola


𝑦2 − 4𝑦 − 8𝑥 − 28 = 0

5. Hyperbolas• The definition of a hyperbola is the same as the definition of an ellipse

except that it is the difference of the two distances that is constant.

• That means:


𝑑1 − 𝑑2 = constant

• Assume that the two foci are placed on the x-axis at (−𝑐, 0) and (𝑐, 0), respectively, and let 2𝑎 be the constant. It follows that

and explicitly:

• Since 𝑎 < 𝑐, we let 𝑏2 = 𝑐2 − 𝑎2. The above expression can be rewritten into the following standard form of a hyperbola:


𝑑1 − 𝑑2 = ±2𝑎

(𝑥 + 𝑐)2 + 𝑦2 − 𝑥 − 𝑐 2 + 𝑦2 = ±2𝑎

𝑥2

𝑎2−𝑦2

𝑏2= 1

• There are some unique features of a hyperbola:

i. Foci: ±𝑐, 0

ii. Vertices: ±𝑎, 0

iii. Co-vertices: 0,±𝑏

iv. Length of transverse axis:

v. Length of conjugate axis:

vi. Asymptotes:

vii. Note that 𝑎 is not necessarily larger than 𝑏!Unit 8 - Conic Sections 41

𝐿𝑡 = 2𝑎

𝐿𝑐 = 2𝑏

𝑦 = ±𝑏

𝑎𝑥


• When the foci of a hyperbola lie on the 𝑦-axis, the hyperbola opens up and down with a vertical transverse axis.

𝑥2

𝑎2−𝑦2

𝑏2= 1

𝑦2

𝑎2−𝑥2

𝑏2= 1

• Example: Find the vertices, foci, and asymptotes of the hyperbola


𝑥2

25−𝑦2

9= 1

• Example: A hyperbola has the equation 9𝑥2 − 16𝑦2 = 144. Find the vertices, foci, and asymptotes. Sketch the graph.


• Example: Find an equation of the hyperbola with vertices 0,−4 , 0, 4and asymptotes 𝑦 = −

𝑥

2, 𝑦 =

𝑥

2.


• Example: Find the equation of the hyperbola with vertices ±3, 0 and foci ±4, 0 . Sketch the graph.


• Applying shifts to hyperbolas leads to the equations and graphs shown below.


(𝑥 − ℎ)2

𝑎2−

𝑦 − 𝑘 2

𝑏2= 1

(𝑦 − 𝑘)2

𝑎2−

𝑥 − ℎ 2

𝑏2= 1

• Example: A hyperbola is described by the equation

a. Identify the center, vertices, and the lengths of transverse and conjugate axes.

b. Determine the equations of the asymptotes and graph the hyperbola.


16 𝑥 + 1 2 − 25 𝑦 − 4 2 = −400

• Example: A shifted conic has the equation

a. Show that the equation represents a hyperbola.

b. Find the center, vertices, foci, and asymptotes of the hyperbola. Sketch its graph.


9𝑥2 − 72𝑥 − 16𝑦2 − 32𝑦 = 16

• Like the ellipse, the equation that defines the eccentricity of a hyperbola is

• Note that 𝑎 < 𝑎2 + 𝑏2, therefore 𝑒 > 1 for hyperbolas.


𝑒 =𝑐

𝑎=

𝑎2 + 𝑏2

𝑎

• Example: Find the eccentricity of the hyperbola.


𝑦2

2−

𝑥 − 1 2

36= 1

6. Review• A short summary – General form


Type of conics

Conditions Discriminant Eccentricity

Circle 𝐵 = 0, 𝐴 = 𝐶 𝐵2 − 4𝐴𝐶 < 0 𝑒 = 0

Ellipse 𝐵 = 0, 𝐴 ≠ 𝐶 𝐵2 − 4𝐴𝐶 < 0 0 < 𝑒 < 1

Parabola 𝐵 = 0, 𝐴 = 0 or 𝐶 = 0 𝐵2 − 4𝐴𝐶 = 0 𝑒 = 1

Hyperbola 𝐵 = 0, 𝐴𝐶 < 0 𝐵2 − 4𝐴𝐶 > 0 𝑒 > 1

𝐴𝑥2 + 𝐵𝑥𝑦 + 𝐶𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0

• Schematically:


• A short summary – Standard form


Type of conics Horizontal Axis Vertical Axis

circle 𝑥 − ℎ 2 + 𝑦 − 𝑘 2 = 𝑟2 𝑥 − ℎ 2 + 𝑦 − 𝑘 2 = 𝑟2

Ellipse (𝑥 − ℎ)2

𝑎2+(𝑦 − 𝑘)2

𝑏2= 1

𝑎 > 𝑏, 𝑐 = 𝑎2 − 𝑏2

(𝑥 − ℎ)2

𝑎2+(𝑦 − 𝑘)2

𝑏2= 1

𝑏 > 𝑎, 𝑐 = 𝑏2 − 𝑎2

Parabola 𝑦 − 𝑘 2 = 4𝑝(𝑥 − ℎ) 𝑥 − ℎ 2 = 4𝑝(𝑦 − 𝑘)

Hyperbola (𝑥 − ℎ)2

𝑎2−(𝑦 − 𝑘)2

𝑏2= 1

𝑎 > 𝑏, 𝑐 = 𝑎2 + 𝑏2

(𝑦 − 𝑘)2

𝑎2−(𝑥 − ℎ)2

𝑏2= 1

𝑎 > 𝑏, 𝑐 = 𝑎2 + 𝑏2

7. Applications of conic sections

• Conic sections have a wide range of applications in science and technology:

i. Astrophysics and Space Science

ii. Optics and Acoustics

iii. Architectures

iv. Communication Technology

v. Navigational Systems

vi. And many more …


A. Parabolas

• Parabolas have a special reflecting property that makes them useful in the design of telescopes, radar equipment, auto headlights, and solar furnaces.

• When a ray of light or sound wave traveling parallel to the axis of symmetry bounces off a parabolic reflecting surface, called paraboloid, it passes through the focus.

Unit 8 - Conic Sections 56Telescope Headlight

• Parabolas are also important in the design of suspension bridges. A support cable in a shape of a parabola can provide the best effect of load bearing of the bridge.


• Example: The Parkes radio telescope has a parabolic dish shape with diameter 210 ft and depth 32 ft. Because of this parabolic shape, distant rays hitting the dish will be reflected directly toward the focus.

a. Determine an equation that models this cross section by placing the vertex at the origin with the parabola opening up.

b. The receiver must be placed at the focus of the parabola. How far from the vertex of the parabolic dish should the receiver be located?


• Example: The cable in the center portion of a bridge is supported to form a parabola. The center vertical cable is 10 ft high, the supports are 210 ft high, and the distance between the two supports is 400 ft. Find the height of the remaining vertical cables, if the vertical cables are evenly spaced. (Ignore the width of the supports and cables.)


• Example: There is a reflector in the Pyrenees Mountains that is eight stories high. It costs $2 million and took 10 years to build. Made of 9000 mirrors arranged in a parabolic formation, it can reach 6000°F just from the Sun hitting it!

• If the diameter of the parabolic mirror is 100 meters and the sunlight is focussed 25 meters from the vertex, find the equation for the parabolic dish.


• Example: Water is flowing from a horizontal pipe 48 ft above the ground. The falling stream of water has the shape of a parabola whose vertex (0, 48) is at the end of the pipe. The stream of water strikes the ocean at the point (10 3, 0). Write an equation for the path of the water.


B. Ellipses

• Ellipses, like parabolas, have an interesting reflection property that if a light source is placed at one focus of the reflecting surface with elliptical cross section, then all the light will be reflected off the surface to the other focus.

• This principle, which works for sound waves and for light, is used in lithotripsy, a treatment for kidney stones.


• The reflection property of ellipses is also used in the construction of whispering galleries. Sound coming from one focus bounces off the walls and ceiling of an elliptical room and passes through the other focus.


• Using his Law of Universal Gravitation, Isaac Newton was the first to prove Kepler’s first law of planetary motion: The motion of each planet about the Sun is an ellipse with the Sun at one focus. (Refer to Physics 12 for the details of the theory of gravitation)


• Example: Suppose that a room is constructed on a flat elliptical base by rotating a semi-ellipse 180° about its major axis. Then, by the reflection property of the ellipse, anything whispered at one focus will be distinctly heard at the other focus. If the height of the room is 16 ft and the length is 40 ft, find the location of the whispering and listening posts.


• Example: The orbit of the planet Mars is an ellipse with the Sun at one focus. The eccentricity of the ellipse is 0.0935, and the closest distance that Mars comes to the Sun is 128.5 million miles. Find the maximum distance of Mars from the Sun.


• Example: The coordinates in miles for the orbit of the artificial satellite Explorer VII can be modeled by the equation

where 𝑎 = 4465 and 𝑏 = 4462. Earth’s center is located at one focus of the elliptical orbit. Find the maximum and minimum heights of the satellite above Earth’s surface to the nearest mile.


𝑥2

𝑎2+𝑦2

𝑏2= 1

• Example: Suppose a lithotripter is based on the ellipse with equation

a. How far from the center of the ellipse must the kidney stone and the source of the beam be placed?

b. What will be their new locations if the equation of the ellipse is 9𝑥2 +4𝑦2 = 36?


𝑥2

36+𝑦2

9= 1

C. Hyperbolas

• The hyperbola has several important applications involving soundingtechniques.

• In particular, the LORAN-C (Long Range Navigation) system was used before 2000’s onboard a ship to determine its location.


• Like the parabola and ellipse, a hyperbola also possess a reflecting property. The Cassegrain reflecting telescope utilizes a convex hyperbolic secondary mirror to reflect a ray of light back through a hole to an eyepiece behind the parabolic objective mirror.


• Another interesting application of hyperbolas involves the orbits of comets in our solar system. Comets can have elliptical, parabolic, or hyperbolic orbits. The center of the Sun is a focus of each of these orbits, and each orbit has a vertex at the point where the comet is closest to the Sun.


• Comets with elliptical orbits, such as Halley’s comet, are the only ones that remain in our solar system. Those with parabolic or hyperbolic orbits, on the other hand, can be seen only once and will never return!

• Example: A hyperbolic mirror (used in some telescopes) has the property that a light ray directed at focus A is reflected to focus B (see figure). Find the vertex of the mirror when its mount at the top edge of the mirror has coordinates (24, 24).


• Example: A cross section of a sculpture can be modeled by a hyperbola.

a. Write an equation that models the curved sides of the sculpture.

b. Each unit in the coordinate plane represents 1 foot. Find the width of the sculpture at a height of 18 feet.


• Example: In the given figure, the LORAN stations A and B are 500 miles apart, and the ship at P receives station A’s signal 2640 𝜇s before it receives the signal from station B.

a. Assuming that radio signals travel at 980 ft/𝜇s, find 𝑑 𝑃, 𝐴 − 𝑑(𝑃, 𝐵).

b. Find an equation for the branch of the hyperbola indicated in red in the figure.

c. If A is due north of B and if P is due east of A, how far is P from A?


• Example: Some comets, such as Halley’s comet, are a permanent part of the solar system, traveling in elliptical orbits around the Sun. Others pass through the solar system only once, following a hyperbolic path with the Sn at a focus. The figure shows the path of such a comet. Find an equation for the path, assuming that the closest the comet comes to the Sun is 2 × 109 miles and that the path the comet was taking before it neared the solar system is at right angle to the path it continues on after leaving the solar system.


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Pre-Calculus 12 - Weebly · Pre-Calculus 12 Unit 8 –Conic Sections. 1. Introduction to conic...

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