To derive an equation for an ellipse, and will represent the foci and for the constants and and . The ellipse is defined by the set of points such that .
Using the distance formula , the length of plus the length of equals will be determined.
The point satisfies this equation if the point is on the ellipse defined by and and . The inverse of this ellipse is one that has the axis as its focal axis and therefore the equation is written as
We now have the standard equation of two conics. The standard equation of the hyperbola is derived by following the process similar to that shown for the ellipse. This is a project that you can complete. Now we look at how these formulae can be used in solving problems.
Example 1: Find an equation in standard form for the parabola that satisfies the following conditions:
a) Focus , directrix
b) Focus and Vertex
Solution:
a) The directrix is and the focus is making the focal length . This means the parabola opens upward. The equation of the parabola in standard form is or .
b) The axis of this parabola is the line passing through the vertex and the focus . The equation of the axis is . Therefore the equation will be of the form where and , so
The equation in standard form is:
Example 2: Find an equation in standard form for the ellipse that satisfies the following conditions:
a) Major axis endpoints (, minor axis length .
b) Major axis endpoints are and ; the minor axis length is .
Solution:
a) The endpoints of the ellipse are on the axis at and . The centre is at the origin . The length of the major axis is so . The length of the minor axis is so . Therefore the equation of the ellipse in standard form is . This makes the equation
b) The endpoints of the ellipse are on the line and the endpoints are located at and . The vertex is located at which is the midpoint of the major axis. The length of the major axis is so and the length of the minor axis is so . The equation of the ellipse in standard form is .
This makes the equation
Prior to converting the standard equation of a conic section to polar form, we must become reacquainted with some terms. Look at the figure below.
is a fixed point known as the focus.
is a point on the fixed line .
The fixed line is called the directrix.
a point on the conic
The ratio of the distance from to and the distance from to is called the eccentricity (e) of the conic. This value will determine the shape of the graph. If the graph will be an ellipse. If , the graph will be a parabola. If , the graph will be a hyperbola.
This equation, can also be written as or
Example 3:
a) Graph the polar equation where and the directrix is . Write the polar equation for the conic and describe the shape of the graph.
b) Graph the polar equation where and the directrix is . Write the polar equation for the conic and describe the shape of the graph.
Solution:
a)
The graph is an ellipse with the axis as its major axis of and the axis as its minor axis. The ellipse has been translated horizontally and is not symmetrical about the pole.
b)
The graph is a parabola opening right with its vertex at and its directrix at .
For a circle that has its center at the origin (pole) the polar form of the equation is and is the radius.
For a circle with radius “a” and passing through the origin the polar form of the equation is or .
Example 4: Graph the equation and describe the graph.
Solution:
The graph is a circle with center and a radius of . The equation in rectangular form would be .
Example 5: Graph the equation of the circle that has a radius of “a” and passes through the origin.
Solution:
In figure 7, the circle passes through the origin and is symmetrical about the positive axis.
In figure 8, the circle passes through the origin and is symmetrical about the negative axis.
In Figure 9, the circle passes through the origin and is symmetrical about the positive axis.
In Figure 10, the circle passes through the origin and is symmetrical about the negative axis.
By changing the value of “a” in the above equations, the axis of symmetry was changed for each circle.
Equations of limaçons have two general forms:
and
The values of “a” and “b” will determine the shape of the graph and whether or not it passes through the origin. When the values of “a” and “b” are equal, the graph will be a rounded heart-shape called a cardiod. The general polar equation of a cardiod can be written as and
Example 6: Graph the following polar equations on the same polar grid and compare the graphs.
Solution:
The cardiod is symmetrical about the positive axis and the point of indentation is at the pole.
The result of changing to is a reflection in the axis.
The cardiod is symmetrical about the negative axis and the point of indentation is at the pole.
Changing the value of “a” to a negative did not change the graph of the cardiod.
Example 7: What affect will changing the values of and have on the cardiod if ? We can discover the answer to this question by plotting the graph of .
Solution:
The cardiod is symmetrical about the positive axis and the point of indentation is pulled away from the pole.
Example 8: What affect will changing the values of and or changing the function have on the cardiod if ? We can discover the answer to this question by plotting the graph of .
Solution:
The cardiod is now a looped limaçon symmetrical about the positive axis. The loop crosses the pole.
The cardiod is now a looped limaçon symmetrical about the positive axis. The loop crosses the pole. Changing the function to cosine rotated the limaçon counter clockwise.
As you have seen from all of the graphs, transformations can be performed on all the rectangular equations as well as the polar equations. The transformations are done by making changes in the constants and/or the functions of the polar equations. Remember the general polar equation for a rose is or . Now you can have some fun and discover the transformations of these graphs by plotting various forms of the equations.
Let’s return to our flashlight. For the light rays to be parallel to the axis of the mirror, the filament of the bulb should be located at the focus. You have learned that the equation of a parabola in standard form . The point that is located on the parabola will be used to determine the value of .
The filament must be placed . from the vertex along the axis of the mirror.
Lesson Summary
In this lesson you learned about the shapes that are classified as conics and how they received the name. You also learned that standard equations of the graphs change when transformations occur. Transformations were then extended to the graphs of polar equations and you learned how to manipulate the equations to produce new images of these shapes.
Points to Consider
Which curves are easiest to represent it with rectangular coordinates and which with polar coordinates?
Is it possible for polar curves to intersect?
Can two different equations produce the same polar curve?
List several ways in which polar representation differs from rectangular representation.
Review Questions
1. Prove that the graph of the equation is a parabola. Determine the vertex, focus and the directrix.
2. Determine the center, vertices, foci and the eccentricity of an ellipse that has as its equation.
3. For the equation , determine the eccentricity, the type of conic and the directrix.
Review Answers
The equation is in standard form . The vertex is and or . Therefore the focus is which equals .
The directrix is or .
Vertex:
Focus:
Directrix:
The equation is in standard form
and the centre
is
. The semimajor axis is
and the vertices are
and
.
and the foci are
and
The eccentricity is
.
Center
Vertices
and
Foci
Eccentricity .66
The numerator and denominator must be divided by 4. If , the graph will be an ellipse. The eccentricity is so the conic is an ellipse. The numerator therefore . The directrix is .
Eccentricity
Conic is an ellipse
Directrix
Applications, Technological Tools
Rectangular Form or Polar Form
Learning Objectives
A student will be able to:
Realize the solutions to real world problems in either rectangular form or polar form.
Manipulate both forms of equations.
Introduction
Sometimes it is not convenient to solve a real world problem using the rectangular coordinates of points nor is it appropriate to express the solution in rectangular form. To simplify the solution and often to create a better overview of the problem, the polar form is more suitable.
1. Mr. Goldbar, the town’s most recent millionaire, wants to erect a large rock-climbing wall in the public park. He feels that this would be entertaining for everyone as well as a great exercise unit for the people. He has access to a flat circular plot of land that has a foot radius. He has marked off a possible location for the wall at coordinates and , where is measured in feet. Sketch the plot of land showing the location of the markers and determine the polar equation of the line between these markers.
Answers for solution:
2. Our local team has qualified for a position in the playoffs. The arena has planned to create a special tribute to the five players who turn and must leave the league. They would like to create a light image at centre ice that will show the faces of the five players on a circle. To do this, they wish to create equal distances between each picture and have it large enough to be seen from the farthest location in the arena. They have a foot circular tube to produce the circle and must work on the location of the pictures. The first photo is placed at and . What is the equation of the line that contains these points and what shape should they create within the circle?
Answers for solution:
The shape that they should create within the circle to enhance the projection at centre ice is a pentagon – one vertex for each picture. The equation of the line containing the first picture is
Polar equations for conics are used extensively when dealing with the orbit of a planet based on the farthest distance from the Sun (aphelion) and the closest distance to the Sun (perihelion). The orbit of a planet is elliptical in shape and each planet has a defined eccentricity and semimajor axis. Using the ellipse shown below and the formula , derive a formula that expresses the standard equation in terms of and . Using this formula, determine the aphelion and perihelion distances of the planet Venus that has a semimajor axis of . and an eccentricity of .
Solution:
Applications, Trigonometric Tools: Polar Coordinates to Rectangular Coordinates
Learning Objectives
A student will be able to:
Use the TI graphing calculator to convert polar coordinates to rectangular coordinates and vice versa.
Introduction
You have learned how to convert back and forth between polar coordinates and rectangular coordinates by using the various formulae presented in this lesson. The TI graphing calculator allows you to use the angle function to convert coordinates quickly from one form to the other. The calculator will provide you with only one pair of polar coordinates for each pair of rectangular coordinates.
Example 1: Express the rectangular coordinates of A as polar coordinates.
Polar coordinates are expressed in the form . An angle can be measured in either degrees or radians, and the calculator will express the result in the form selected in the [MODE] menu of the calculator.
Press [MODE] and cursor down to Radian Degree. Highlight radian. Press [ MODE] to return to home screen. To access the angle menu of the calculator press [ APPS] and this screen will appear:
Cursor down to and press [ENTER] or press on the calculator. The following screen will appear . Press ) [ENTER] and the value of r will appear . Press [CLEAR]. Access the angle menu again by pressing [ APPS]. When the angle menu screen appears, cursor down to and pres [ENTER] or press on the calculator. The screen will appear. Press ) [ENTER] and the value of will appear .
This procedure can be repeated to determine the rectangular coordinates in degrees. Before starting, press [MODE] and cursor down to Radian Degree and highlight degree.
Example 2: Express the polar coordinates of in rectangular form.
The angle is given in degrees so the mode menu of the calculator should also be set in degree. Therefore, press [MODE] and cursor down to Radian Degree and highlight degree. Press [ MODE] to return to home screen. To access the angle menu of the calculator press [ APPS] and this screen will appear:
Cursor down to and press [ENTER] or press on the calculator. The following screen will screen will appear: Press 300, 70) and the value of will appear Press [CLEAR]. Access the angle menu again by pressing [ APPS]. When the angle menu screen appears, cursor down to and pres [ENTER] or press on the calculator. The screen will appear. Press ) [ENTER] and the value of y will appear .
Applications, Trigonometric Tools: Graphs of Polar Equations
Learning Objectives
A student will be able to:
Use Geometer’s Sketchpad software to display the graph of a polar equation.
Introduction
A graphing calculator is a very good source of technology for students. It is compact, portable and readily accessible. However, most students also have access to a computer. This software would be an asset for any student and it presents visual representations that are larger than those displayed on a calculator screen. The process involved in producing the graph acts as a valuable learning tool for the student. In this lesson, the students will learn how to graph a polar equation using Geometer’s Sketchpad.
The software program, Geometer’s sketchpad, is extremely useful in graphing polar coordinates and polar equations. We will go through the process of graphing the polar equation .
To begin, left click on Graph. Scroll down to Grid Form and over to Polar Grid and left click. The following screen appears:
This screen may be maximized like any document. If the grid seems off centre, point and click on the red dot of the origin and drag the grid to where you want it on the screen. The tool at the top of the upper left corner should be highlighted. This is the arrow and it is the select tool. Also, the red dot – the unit point on the axis can be hidden. Point on the dot and right click. A list of options will appear. Scroll down and highlight Hide Unit Point. Left click and the point disappears. To rescale the graph, left click on a number on either axis until a double arrow appears. Drag the number toward the origin until the proper scale is reached. Notice the difference in the scale of this figure and the previous one.
To enter the equation, left click on Graph and scroll down to New Function. The equation editor appears.
Enter the equation: Left click on equation and scroll down to . Using the keypad, left click on functions and scroll to and then . This function will appear in the upper left corner of the grid.
To plot the New Function, left click on Graph and scroll down to Plot New Function. A screen may appear asking you if you want the graph in radians. Click yes and the graph will appear on the grid. The graph will be a fuzzy pink picture. Point and click on the plotted graph and it will be restored to a smooth line.
To copy the graph, le
ft click on Edit and scroll down to Select All. This will highlight the entire page. Then left click on Edit again and highlight Copy. You can now paste the graph in a document or print it. The appearance of the graph can be changed by using the Display menu. Left click on Display and scroll to line width. You can select dashed, thin or thick. In the same menu, you can change the color of the graph.
Graph and Calculate Intersections of Polar Curves
Given Two Polar Curves, Find All Intersection Points
Learning Objectives
A student will be able to:
Graph polar curves to see the points of intersection of the curves.
Understand the difficulty of determining polar coordinates for intersection points.
Use Cartesian coordinates to determine points of intersection.
Introduction
Josie has painted two murals that she is trying to combine to form one large mural. The mural is going to hang on a wall in the front entrance of her new home. After several attempts, Josie has decided that she should overlap the paintings to produce the most appealing view of her art. If she does this, where will the murals intersect?
CK-12 Trigonometry Page 28
