UNIT 15 Conic Sections The telescopes at the W.M. Keck Observator in Hawaii use hperbolic mirrors. Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Astronomers use telescopes with optics that use the properties of conic sections. Cassegrain reflectors, for eample, consist of two conic-shaped mirrors. Big Ideas Conic sections have useful applications in the phsical world for eample, in antenna design, optics, architecture, and structural engineering. If ou can create a mathematical model for a situation, ou can use the model to solve other problems that ou might not be able to solve otherwise. Unit Topics Foundations for Unit 15 Introduction to Conic Sections Circles Ellipses Hperbolas Parabolas Putting Conics into Graphing Form CONIC SECTIONS 01 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Foundations for Unit 15 Before ou learn more about conic sections, make sure ou know how to do the following: Use the geometric definition of a circle and its parts. Find coordinates on a plane when given a description. Use square roots to solve simple equations. Using the Geometric Definition of a Circle and Its Parts Definitions circle the set of all points in a plane that are the same distance r from a given point in the plane radius a line segment that connects the center of a circle to a point on the circle chord a line segment that connects two points on a circle diameter a line segment that connects two points on a circle and contains the center of the circle; the length of the diameter d is two times the length of a radius r : d = r or r = d Eample 1 Find the radius r and diameter d of each circle. A B radius diameter center chord 3 units 8 units The radius is 3 units long, so r = 3. The diameter is 8 units long, so d = 8. The diameter is twice the radius, so The radius is half the diameter, so d =. r =. 0 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Unit 15 Foundations Problem Set A Find the radius r and diameter d of each circle. 1. 3... Using Smmetr to Find Coordinates of a Point A figure has line smmetr if there is a line on which the figure ma be folded so that two parts of the figure are mirror images of each other. SPECIAL LINES OF SYMMETRY Line of Smmetr -ais Smmetric Points (, ) is smmetric to (, ) -ais (, ) is smmetric to (, ) = (, ) is smmetric to (, ) = (, ) is smmetric to (, ) If point A is smmetric to point A with respect to line m, the midpoint AA is point M, the intersection of AA and line m. Therefore, AM = MA. of FOUNDATIONS FOR UNIT 15 03 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Unit 15 Foundations Eample For each of the following, A is smmetric to A with respect to the given line. Find the coordinates of point A. A A(, ), line of smmetr: -ais (, ) is smmetric to (, ) with respect to the -ais. So A(, ) is smmetric to A (, ). B A( 1, 5), line of smmetr: = (, ) is smmetric to (, ) with respect to the line =. So A( 1, 5) is smmetric to A (5, 1). C A(3, 1) is smmetric to A with respect to the line =. Create a graph. 3 A (3, 1) A' (5, 1) 1 1 unit 1 unit 1 1 1 3 5 3 Problem Set B Point A is located at (5, 1). For each of the following, A is smmetric to A with respect to the given line. Find the coordinates of point A. 5. A(5, 3), line of smmetr: -ais 8. A(, ), line of smmetr: =. A(, ), line of smmetr: -ais 9. A(, ), line of smmetr: = 5 7. A(7, 3), line of smmetr: =. A(, 1), line of smmetr: = 3 Solving Equations of the Form = d SQUARE ROOT PROPERTY If d > 0, then = d has two real number solutions, = d and = d. Eample 3 Solve the equation. Put the result in simplified radical form. A n = 0 B z = 50 n = ± 0 z = ± 50 n = ± 5 z = ± 5 n = ± 5 z = ±5 THINK ABOUT IT The square root propert is sometimes written, If = d, then = ± d. Problem Set C Solve each equation. Put the result in simplified radical form. 11. 1 1. = 7 1. = 81 15. 1 13. = 35 1. 38 17. = 57 18. 0 19. = 3 0 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Introduction to Conic Sections As the name suggests, a conic section is a cross-section of a cone. Identifing Conic Sections DEFINITIONS A conic section (or conic) is a two-dimensional graph that can be formed b the intersection of a plane with a double-napped cone. A double-napped cone is formed as follows: One line, called the generating line, intersects and revolves around another line, called the ais. The ais is stationar, and the two lines cannot be perpendicular. The point where the lines intersect is the verte of the cone. ais generating line nappes verte THINK ABOUT IT A nappe is one of two equal pieces of a cone when the cone is divided at the verte b a plane perpendicular to the ais. Note: The double-napped cone described above is a surface without an bases. If a circular base were added to one nappe, the resulting figure would be the familiar cone that ou stud in geometr. INTRODUCTION TO CONIC SECTIONS 05 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
There are four important tpes of conic sections. Circle Ellipse Parabola Hperbola A circle is a conic section formed when a plane intersects onl one nappe, perpendicular to the ais. An ellipse is a conic section formed when a plane intersects onl one nappe, not parallel to the generating line. A circle can be considered a special kind of ellipse. A parabola is a conic section formed when a plane intersects onl one nappe, parallel to the generating line. A hperbola is a conic section formed when a plane intersects bothnappes. There are also three tpes of degenerate conic sections. Degenerate Conic Sections A degenerate conic section is formed when a plane intersects the verte of the double-napped cone. There are three degenerate cases. Point The plane intersects onl the verte. This is a degenerate ellipse. Line The plane contains the generating line. This is a degenerate parabola. Pair of Intersecting Lines The plane intersects both nappes through the verte. This is a degenerate hperbola. TIP In mathematics, degenerate cases are cases that are simpler than normal cases. Points and lines are degenerate conic sections because the are simpler than the other conic sections. 0 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Eample 1 Identif the conic that will be formed b the intersection of the plane and the cone. A Given: Plane A intersects line m, is not perpendicular to m, and is not parallel to n. The conic is an ellipse. m (ais) A n (generating line) REMEMBER A line etends without end in one dimension. A plane etends without end in two dimensions. B Given: Plane B is parallel to line n. The conic is a parabola. C Given: Plane C is perpendicular to line m and at the center of the cone. The conic is a degenerate ellipse (a point). C B Using Tools of Analtic Geometr: The Distance Formula Analtic geometr (also called coordinate geometr) is the stud of geometr using the tools of algebra. You ll use the tools of analtic geometr to stud conic sections. The distance formula is one important tool in analtic geometr. The distance formula is based on the Pthagorean theorem. To find an B (, ) epression for the distance d from d A to B, form right triangle ABC and use the Pthagorean theorem. (AB) = (AC ) + (BC ) d = ( 1 ) + ( 1 ) d = ( 1 ) + ( 1 ) DISTANCE FORMULA A ( 1, 1 ) The distance d between two points ( 1, 1 ) and (, ) is d = ( 1 ) + ( 1 ). C (, 1 ) Eample Find the length of AB. Solution Let A(1, ) be ( 1, 1 ) and B(, ) be (, ). AB = ( 1 ) + ( 1 ) = ( 1) + ( ) = 5 + ( ) = 5 + = 9 7 5 3 1 A(1, ) B(, ) 1 1 3 5 7 1 REMEMBER The length of AB is denoted AB. INTRODUCTION TO CONIC SECTIONS 07 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Using Tools of Analtic Geometr: The Midpoint Formula The midpoint formula, another useful tool in analtic geometr, helps ou determine the coordinates of the midpoint of a segment. MIDPOINT FORMULA The coordinates of the midpoint M of the segment with endpoints ( 1, 1 ) and (, ) is M = ( + 1 1 +, ). Eample 3 A Find the coordinates of M, the midpoint of AB in Eample. Solution ( + 1 1 +, ) = ( 1 + +, ) The coordinates of M are (3.5, 3). B A line segment CD has one endpoint D at (, ) and a midpoint M at (0,.5). Find the coordinates of the other endpoint C. THINK ABOUT IT You can use the distance formula to verif that M is the midpoint. Show that AM = MB. AM = (3.5 1) + (3 ) = 7.5 MB = ( 3.5) + ( 3) = 7.5 5 3 1 3 1 1 D(, ) M(0,.5) 1 3 Solution ( + +, ) = (0,.5) + + = 0 =.5 + = 0 + = 5 = Substitute the coordinates of the given endpoint (, ) and the midpoint (0,.5) into the formula. Set each epression equal to its corresponding coordinate, and then solve for and. The coordinates of C are (, 1). 08 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Problem Set Identif the conic that will be formed b the intersection of the plane described and the cone below. m (ais) 1. Given: The plane is perpendicular to m and above the verte.. Given: The plane intersects m and n at the same point. 3. Given: The plane contains n.. Given: The plane is parallel to n and intersects one nappe. 5. Given: The plane intersects onl one nappe and is neither perpendicular to m nor parallel to n.. Given: The plane intersects both nappes. 7. Given: The plane intersects both nappes through the verte. Find the eact length of AB. Epress our answer in simplified radical form. n (generating line) 8. A(1, 5), B( 3, ) 9. A(, ), B(, ). A(, ), B(, 9) 11. A(, 9), B(7, 5) 1. A(3, ), B(1, 8) 13. A(3, 9), B(8, ) 1. A( 9, 8), B( 5, 8) 15. A(, ), B(, ) * 1. Challenge A(., 5.3), B(.5, ) Find the coordinates of midpoint M of AB. 17. A(, 8), B(, 9) 18. A(, ), B(8, 7) 19. A( 3, ), B(, 3) 0. A(, ), B(, ) 1. A(1, ), B(1, ). A(, 7), B(, ) Find the coordinates of endpoint B of AB, given midpoint M and endpoint A. 3. A(, 0), M( 1, 0). A(, 3), M( 7, 3) 5. A(, 1), M(1, 0). A(, ), M( 3, 1) 7. A( 3, 1), M( 5, 5) 8. A(, ), M(0, ) Solve. * 9. Challenge A right triangle has vertices A( 5, ), B( 3, 8), and C( 3, ). The triangle is then rotated 180 to the right. Find the coordinates of the vertices of the new triangle b using the midpoint formula. INTRODUCTION TO CONIC SECTIONS 09 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Circles The circle is the simplest and most familiar conic section. Deriving an Equation of a Circle from the Definition DEFINITION A circle is the set of all points in a plane that are a fied distance r (the radius) from a given point (the center). In a coordinate plane, a circle is a set of points (, ). For an circle, r has the same value, no matter where (, ) is on the circle. r center (, ) TIP The word radius can mean either a line segment from the center to the circle, or the length of such a line segment. Eample 1 Derive an equation of the circle with the given center and radius. A center = (0, 0), r = Solution For an point (, ) on the circle, the distance from the center (0, 0) is. ( 1 ) + ( 1 ) = d Distance Formula ( 0) + ( 0) = The distance between (0, 0) and an point (, ) on the circle is. (0, 0) (, ) ( 0) + ( 0) = 3 Square each side. + = 3 Simplif. B center = (, 5), r = 3 Solution For an point (, ) on the circle, the distance from the center (, 5) is 3. ( ) + ( 5) = 3 The distance between (, 5) and an point (, ) on the circle is 3. ( ) + ( 5) = 9 Square each side. (, 5) 3 (, ) UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Writing an Equation of a Circle, Given Its Graph There are different forms for the equation of a circle, but one common form is a helpful one to use when ou want to graph the circle. (, ) EQUATION OF A CIRCLE The equation in graphing form of the circle with center (h, k) and radius r is ( h) + ( k) = r. (h, k) r Eample Write the equation in graphing form of the circle shown at right. Solution The center is ( 3, ), so h = 3 and k =. To find the radius, calculate the distance between the center ( 3, ) and the point (0, ) on the circle: center ( 3, ) 5 (0, ) 5 r = (0 ( 3)) + ( ) = 9 + 1 = 5 = 5. Substitute h, k, and r into the equation in graphing form: ( h) + ( k) = r ( ( 3)) + ( ) = 5 ( + 3) + ( ) = 5 Finding the Center, Radius, and Diameter of a Circle, Given Its Equation in Graphing Form Eample 3 Find the center, radius, and diameter of the circle with equation ( + 1) + ( + 3). Solution Rewrite the equation as ( ( 1)) + ( ( 3)) = to identif h, k, and r. The circle is in graphing form with h = 1, k = 3, and r =. The center of the circle (h, k) is ( 1, 3). The radius r is, so the diameter is = 8. CIRCLES 11 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Graphing a Circle, Given Its Equation in Graphing Form Eample Graph the circle with equation ( ) + ( + 1) =. Solution The equation is in graphing form with h =, k = 1, and r =. Step 1 Plot the center (h, k) at (, 1). Step The radius is, so ever point on the circle is units from the center. Count units left, right, up, and down from the center to find four points on the circle: (, 1), (, 1), (, 1), and (, 3). Step 3 Draw the circle through the four points. 5 (, 1) (, 1) 5 (, 1) (, 3) (, 1) TIP When ou draw a circle, use a compass to improve our accurac. Finding the Equation of a Circle, Given Three Points Eample 5 Find the equation of the circle that contains A( 3, ), B(3, ), and C(5, 0). Solution You need the center and radius. To find the center, find the point of intersection of the perpendicular bisectors of two chords of the circle. Step 1 Identif two chords, sa AB and BC. Find the slope of each chord. Then find the slope of each perpendicular bisector b using the opposite reciprocal. Chord Slope of Chord Slope of Perpendicular Bisector AB BC _ 3 ( 3) = = 3 1 3 0 5 3 = = 1 Step Find the midpoint of each chord. Midpoint of AB : ( 3 + 3 +, ) = ( 0, ) = (0, 5) Midpoint of BC : ( 3 + 5, + 0 ) = ( 8, ) = (, ) Step 3 Find the equations of the perpendicular bisectors. Use the point-slope form 1 = m( 1 ). Use the slopes from Step 1 and the midpoints from Step. Perpendicular bisector of AB : 5 = 3( 0), or = 3 + 5. Perpendicular bisector of BC : = 1 ( ), or = 1. REMEMBER A chord is a segment that connects two points of a circle. A theorem of geometr states that the perpendicular bisector of an chord of a circle passes through the center of the circle. TIP This problem does not require a graph, but a sketch helps ou visualize what is happening. = 3 + 5 A( 3, ) = 1 B(3, ) (0, 5) (, ) C(5, 0) (, 1) ( + ) + ( + 1) = 50 1 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Step Solve the sstem of linear equations. = 3 + 5 Perpendicular bisector of AB 1 = 3 + 5 Substitute 1 for. = Solve for. ( ) = 1 Substitute and solve for. So the solution to the sstem is (, 1). This is the center of the circle, so (h, k) = (, 1). Step 5 Calculate the radius. Use the center (, 1) and an point on the circle, sa A( 3, ). r = ( 3 ( )) + ( ( 1)) = 1 + 9 = 50 Step Use the center and radius to write the equation in graphing form. ( h) + ( k) = r ( ( )) + ( ( 1)) = ( 50 ) ( + ) + ( + 1) = 50 Problem Set Find the equation of each circle. 1. center at (3, 1) and radius 5. center at ( 5, ) and radius 3. 8 (5, ) (8, ) 8. 5 3 1 1 1 3 5 7 8 9 11 1 1 (8, 1) (, 1) 3 5 5. center at (, 7) and radius 3.5. center at (7, 9) and radius CIRCLES 13 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
7. 8. ( 3, ) 8 7 5 3 1 8 1 ( 3, ) 3 5 9 3 15 1 9 3 3 ( 13, 3) 3 ( 3, 3) 9 1 15 9 For each problem, do the following: A. Find the center, radius, and diameter of the circle. B. Graph the circle. 9. ( ) + ( ). ( ) + ( + 8) = 11. ( + ) + ( + 9) 00 1. ( 1) + ( ) = 5 13. + ( + ) = 3 1. ( 7) + ( + ) = 3 15. + = 9 1. ( 11) = 50 17. ( 1) + ( ) = 18. ( + 8) + ( 5) 19. ( + 5) + ( + 7) = 81 0. ( 1) + Find the equation of the circle that contains the given points. 1. A(, 9), B(1, 1), C(8, 7) 3. A(0, 0), B( 1, 1), C( 1, 3). A(, 3), B( 1, ), C(, 3). A( 3, ), B(9, 0), C(1, 8) 1 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Solve. 5. A satellite travels in a circular orbit around the earth at a distance of,199 miles from the earth s center. A. Assume that the earth s center is positioned at (0, 0) in the solar sstem. Write an equation for the path of the satellite s orbit around the earth. B. Write an equation of the path of the satellite if the earth s center is positioned at ( 000, 1300) in the solar sstem.. An airport s luggage carousel moves in a circular path, as shown in the graph. 5 3 1 1 1 3 5 7 8 9 1 3 5 A. Write an equation for the path of the luggage carousel. B. The luggage comes up in the center of the circle and moves to the circular path in a straight line. What is the distance a piece of luggage travels before it hits the circular path? 7. A ride in an amusement park takes people in a circular loop represented b the equation ( + 3) + ( + ) =, with the center of the park at (0, 0). A. A waterfall is located at the center of the ride. What are the coordinates of the waterfall s location in the amusement park? B. What is the distance between the center of the ride and the center of the park? 8. A radio signal reaches all locations within a 0-mile distance from the radio tower, located at ( 150, 50). A. Write an equation that represents the locations within a 0-mile distance from the radio tower. B. A house is located at (, 0). How far awa is the house from the radio tower? Will the radio signal reach the house? * 9. Challenge Write the equation of a circle with a circumference of 1π and center ( 3, 8). * 30. Challenge A line is tangent to a circle if it intersects the circle in eactl one point. Write the equation of the line that is tangent to the circle ( + 1) + ( + ) = 50 at the point (, 1). CIRCLES 15 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Ellipses When the general second-degree equation A + B + C + D + E + F = 0 has the same signs for A and C (A 0 and C 0), the equation is that of an ellipse. In this topic, ou ll work onl with ellipses that have B = 0. Deriving an Equation of an Ellipse from the Definition DEFINITION TIP An ellipse is the set of all points in a plane such that the sum of the distances from two fied points (the foci) is constant. focus (, ) d 1 d focus Focus (fō-kә s) is singular. Foci (fō-sī) is plural. Verte (vә r-teks) is singular. Vertices (vә rt-ә -sēz) is plural. d 1 + d is constant, no matter where (, ) is on the ellipse. The major ais of an ellipse is the segment that passes through the foci, with endpoints on the ellipse; those endpoints are called vertices. The minor ais is the perpendicular bisector of the major ais; its endpoints are also on the ellipse and are called covertices. The intersection of the aes is the center of the ellipse. The center is the midpoint of both aes: Each half of the major ais is a semimajor ais; each half of the minor ais is a semiminor ais. The major ais is the longer ais. verte coverte minor ais major ais focus center focus coverte verte 1 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Below, F 1 and F are foci, V 1 and V are vertices, V 1 V is the major ais, C 1 and C are covertices, C 1 C is the minor ais, and (h, k) is the center. Ellipse with horizontal major ais: Ellipse with vertical major ais: V C V 1 V F 1 (h, k) F c (, ) b } C 1 a F a c C 1 (h, k) b (, ) F 1 C Equation in Graphing Form: ( h) + ( k) a b V 1 Equation in Graphing Form: ( h) + ( k) b a PROPERTIES OF ELLIPSES Length of major ais (V 1 V in the diagrams) = a Length of semimajor ais (center to either verte) = a Length of minor ais (C 1 C in the diagrams) = b Length of semiminor ais (center to either coverte) = b Distance from center to either focus = c c = a b (or a = b + c ) a > b (But a ma be in either denominator.) Sum of distances from the two foci to an point on the ellipse: d 1 + d = a. TIP A circle can be thought of as a special case of an ellipse, with a = b. In that case, both foci coincide with the center. The ellipse at right has foci (, 0) and (, 0) and constant sum. The equation in graphing form of this ellipse can be derived b using the definition of an ellipse. For an point (, ) on the ellipse, the sum of the distances from the foci is. (, ) d 1 d F 1 (, 0) F (, 0) ELLIPSES 17 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
d 1 + d 0 The sum of the distances is constant. ( + ) + ( 0) + ( ) + ( 0) 0 Distance Formula + 8 + 1 + + 8 + 1 + 0 + 8 + 1 + 0 8 + 1 + + 8 + 1 + 00 0 8 + 1 + + 8 + 1 + 1 0 = 0 8 + 1 + 5 300 +,000 = 00( 8 + 1 + ) Epand binomials. Isolate one radical. Square both sides. Isolate the remaining radical. Square both sides. 5 300 +,000 = 00 300 + 00 + 00 Distribute. 1 00 = 300 Isolate the constant. 5 + 9 Divide both sides b 300. Finding the Length of the Major and Minor Aes and the Center of an Ellipse, Given Its Graph Eample 1 following: Given the ellipse in the graph at right, find the A the lengths of the major and minor aes B the coordinates of the center Solution A The endpoints of the major ais are ( 5, 9) and ( 5, 5), so its length is 9 ( 5). The length of the semimajor ais is a = 7. The endpoints of the minor ais are ( 7, ) and ( 3, ), so its length is 3 ( 7) =. The length of the semiminor ais is b =. B The center (h, k) is the midpoint of either ais: (h, k) = ( 5 5 5 + 9, ) = ( 5, ) or (h, k) = ( 7 3 +, ) = ( 5, ). verte ( 5, 9) 5 coverte coverte ( 7, ) ( 3, ) 5 5 verte ( 5, 5) 18 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Writing an Equation of an Ellipse, Given Its Graph Eample Write the equation in graphing form of the ellipse shown. Solution The values of h, k, a, and b are needed for the equation. The center is the midpoint of the minor ais connecting the covertices, so the center is (h, k) = (3, ). The length of the semiminor ais is b = 5, so b = 5. The length of the semimajor ais is a = 8, so a =. Substitute a, b, h, and k into the equation in graphing form. ( h) The major ais is vertical, so use The equation is ( 3) 5 + ( + ) + ( k) b a.. (3, ) 8 8 (, ) 8 1 (8, ) 1 (3, 1) Graphing an Ellipse, Given Its Equation ( 1) Eample 3 Graph the ellipse with equation 81 + ( + 7). 1 Solution In an ellipse, a > b, so a = 81, and the equation is in graphing ( h) form + ( k). Therefore, the major ais is horizontal, a b h, k = 7, a = 81 = 9, and b = 1 =. Step 1 Plot the center (h, k) at (1, 7). Step The major ais is horizontal. Count 9 units left and 9 units right from the center. Plot the vertices at ( 8, 7) and (, 7). Step 3 The minor ais is vertical. Count units down and units up from the center. Plot the covertices at (1, 11) and (1, 3). Step Sketch the ellipse through the vertices and covertices. (1, 3) ( 8, 7) (, 7) (1, 7) (1, 11) 15 ELLIPSES 19 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Finding the Eccentricit and Foci of an Ellipse All conic sections have a ratio e, called eccentricit. For ellipses, the value of e determines the shape of the ellipse. ( You will see later that the value of e also determines the shape of a hperbola.) DEFINITION The eccentricit of an ellipse is the ratio e = c a. Properties of Eccentricit for Ellipses 0 < e < 1 (or 0 e < 1 if a circle is considered to be an ellipse.) If e is near 0, then c is near 0, and the ellipse looks nearl circular. (The foci are near the center.) If e is near 1, then c and a are close in value, and the ellipse looks elongated. (The foci get closer to the vertices.) c = 0 a V 1 center F 1 and F V V 1 V V 1 F center F F 1 F 1 center a a c c V Eccentricit is 0. Ellipse is a circle. Eccentricit is closer to 0. Ellipse is more circular. Eccentricit is closer to 1. Ellipse is less circular. ( + 3) Eample For the equation 5 + ( 5), do the following: A Find the coordinates of the foci. B Find the eccentricit. Solution The equation is in graphing form with a horizontal major ais. ( h) A Calculate c to find the coordinates of the foci: c = a b = 5, so c = 1. + ( k) a b, The distance from the center to each focus is 1. The major ais is horizontal, so the foci are left and right of the center at ( 3 1, 5) and ( 3 + 1, 5), or about ( 7., 5) and (1., 5). B Calculate a: a = 5 = 5 Substitute the values for c and a: e = a c = 1 5 0.9. THINK ABOUT IT For an ellipse, c = a b. If a = b, then the major ais and minor ais have equal lengths, forming a circle. Then c = 0, c = 0, and e = a c = a 0 = 0. 0 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Problem Set Find the equation of the ellipse. 1.. 8 8 (1, 3) ( 1, 8) 8 C(1, 8) (3, 8) 1 1 (1, 13) 1 (0, ) (, ) 8 (, ) 8 8 (0, ). ( 3, 7) 8 5. (, 7) (, ) C( 3, ) (, ) ( 5, 3) 5 (9, 3) 1 8 ( 3, 3) 8 5 (, 1) 5. 3. (, 1) 1 1 8 ( 1, 5) ( 8, 5) 8 (, 11) 1 18 (, 1) 1 1 1 8 (, 7) (, 7) 8 8 1 (, ) ELLIPSES 1 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
For each problem, do the following: A. Find the coordinates of the center. B. Find the length of the major ais. C. Find the length of the minor ais. D. Graph the ellipse. 7. + 9 13. ( ) 19 + ( + 3) 5 8. + 9 5 1. ( + 7) 81 + _ ( + ) 1 9. ( 1) 1 + 3 15. ( 1) 1 + ( 8) 9. 1 + ( 9) 1. ( 5) 0 + ( ) 81 11. ( 3) 5 + ( ) 17. ( + 1) 5 + ( ) 1 1. ( + 1) _ 0 + _ ( 15) 18. ( 8) 0 + _ ( + ) Find the coordinates of the foci of the ellipse. 19. ( ) 0 + ( 8). ( + ) _ 9 + 0. ( + ) + ( 9) 50 3. ( ) 9 + ( 3) 5 1. ( + ) + ( + 8) 1. ( + 5) + ( ) 9 Find the eccentricit of the ellipse. 5. ( + ) 0 + ( 7) 81 7. ( 9) + ( + 8) 1. 8. 1 (0, ) (, ) 8 (, ) 8 8 (0, ) ( 3, 7) 8 (, ) (, ) 1 8 ( 3, 3) UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Solve. 9. An elliptical mirror reflects light from one focus toward the other focus. A designer installs an elliptical mirror modeled b the equation ( + ) 1 + 5. A. Graph the elliptical mirror. B. A light ra that passes through one focus will pass through the other. What two points will this light ra pass through? 30. A tunnel has an elliptical opening that is 0 feet high and 80 feet wide. A. Write an equation that models the tunnel. B. Find the foci of the ellipse. C. Find the eccentricit of the ellipse. What does this indicate about the shape of the ellipse? 31. A whispering galler is enclosed in an elliptical area. If a person stands at one focus and whispers, the sound will reflect directl to the other focus. Emma and Jamal are in an elliptical whispering galler that is 3 feet long and 19 feet wide. A. What are the lengths of the semimajor and semiminor aes? B. Write the distance that Emma should stand from Jamal so that she will hear Jamal s whisper. C. Write an equation of an ellipse that models the whispering galler. 3. A standing tin of chocolates has an elliptical opening that is 3 centimeters wide and 1 centimeters high. A. Write an equation that models the tin. B. Find the foci of the ellipse. C. Find the eccentricit of the ellipse. What does this eccentricit indicate about the shape of the ellipse? * 33. Challenge The earth is at one focus of the elliptical orbit of a satellite. The length of the major ais is 30,000 miles, and the earth s center is 9000 miles from the center of the ellipse. Write the equation for the path of the satellite. * 3. Challenge The whisper chamber in the United States Capitol is shaped like an ellipse. If Yasmine stands at one focus and whispers, Sharika, who is standing at the other focus, can hear her distinctl. Find the equation of the shape of the room given the length of the horizontal major ais of the chamber is 9 feet and the two girls are standing 71 feet apart. Assume the chamber is centered at the origin. ELLIPSES 3 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Hperbolas When the general second-degree equation A + B + C + D + E + F = 0 has opposite signs for A and C, then the equation is that of a hperbola. Deriving an Equation of a Hperbola from the Definition DEFINITION A hperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fied points (the foci) is constant. d 1 d (, ) Ever hperbola has two separate branches. The line through the foci intersects the hperbola at two vertices. The transverse ais is the line segment joining the vertices. The midpoint of the transverse ais is the center of the hperbola. Ever hperbola has an associated rectangle. The diagonals of the associated rectangle pass through the center. The lines containing these diagonals are asmptotes of the hperbola. As a point in the hperbola gets farther from either verte, it approaches but never touches one of the asmptotes. The conjugate ais is the line segment through the center, perpendicular to the transverse ais, with endpoints on the associated rectangle. A hperbola is smmetric with respect to both its transverse ais and its conjugate ais. focus focus d d 1 is constant, no matter where (, ) is on the hperbola. UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
In each diagram below, F 1 and F are foci, V 1 and V are vertices, V 1 V is the transverse ais, and (h, k) is the center. Hperbola with horizontal transverse ais: Hperbola with vertical transverse ais: = b a = b a (, ) P a b a F 1 V 1 c F 1 V V 1 F c (h, k) P b (h, k) Q V Q (, ) = a b F a = b Equation in Graphing Form: ( h) ( k) a b Slopes of Asmptotes: ± a b Equation in Graphing Form: ( k) ( h) a b Slopes of Asmptotes: ± a b PROPERTIES OF HYPERBOLAS Length of transverse ais = a (V 1 V in the diagrams above) Distance from center to either verte = a Length of conjugate ais = b (PQ in the diagrams above) Distance from center to either focus = c c = a + b In the equation in graphing form, a is alwas the first denominator. An of these are possible: a > b, a = b, or a < b. The hperbola to the right has foci (0, 5) and (0, 5) and constant difference. The equation in graphing form of this hperbola can be derived b using the definition of a hperbola. For an point (, ) on the hperbola, the absolute value of the difference of the distances from the foci is. focus 5 (0, 5) 5 5 5 focus (0, 5) HYPERBOLAS 5 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
d d 1 = ± ( 0) + ( + 5) ( 0) + ( 5) = ± + + + 5 + + 5 = ± + + + 5 = ± + + + 5 + + + 5 = 3 ± 1 + + 5 + + + 5 0 3 = ±1 + + 5 5 9 = ±3 + + 5 5 90 + 81 = 9( + + 5) 5 90 + 81 = 9 + 9 90 + 5 1 9 9 1 If d d 1 =, then d d 1 = ±. Distance Formula Epand binomials. Get one radical on each side. Square both sides. Isolate the radical term. Divide both sides b. Square both sides. Distribute. Isolate the constant term. Divide both sides b 1. Finding Everthing You Want to Know About a Hperbola, Given Its Equation Eample 1 The equation of a hperbola is 9 1. A Identif the location of the center. B Identif the lengths of the transverse and conjugate aes. C Identif the equations of the asmptotes. D Graph the hperbola. Label the aes and asmptotes. Solution Rewrite the equation as ( 0) ( 0) to see that it is in the form ( k) ( h). 3 a b This hperbola has a vertical transverse ais. A In the equation, the center (h, k) is (0, 0). B In the equation, a = 3 and b =. This means the length of the transverse ais is a = 3 =, and the length of the conjugate ais is b = = 8. C The slopes of the asmptotes are ± a b = ± 3, so the equations of the asmptotes are = 3 and = 3. D = 3 transverse = 3 (0, 0) 8 conjugate 8 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Writing an Equation of a Hperbola, Given Its Graph Eample Write the standard equation of the hperbola shown at the right. Solution The center is the midpoint of the line segment connecting the vertices, so the center is (3, ), and h = 3 and k =. The distance between the vertices is, so a = 3. Since the length of the conjugate ais b is, b is 5. The transverse ais is horizontal, so the equation has the form ( h) a ( k) b. The equation is ( 3) ( ). 9 5 8 (0, ) (3, 7) (, ) 8 1 (3, 3) 8 Graphing a Hperbola, Given Its Equation Eample 3 Graph the hperbola with equation ( 3) ( + ). 9 Solution The equation is in the graphing form ( k) ( h), a b so the transverse ais is vertical. In the equation, k = 3 and h =, so the center (h, k) is (, 3). Also, a = and b = 9, so a = and b = 3. Step 1 Graph the center. To graph the vertices, count units up and units down from the center. The vertices are (, 5) and (, 1). Step Use the values of a and b to sketch the associated rectangle. The dimensions of the rectangle are a = and b =. Step 3 Sketch the lines containing the diagonals of the rectangle; these lines will be the asmptotes of the hperbola. 3 3 (, 5) (, 3) (, 1) Verif that the slopes of the asmptotes are ± a b = ± 3. Step Sketch the two branches of the hperbola, using the asmptotes as a guide. HYPERBOLAS 7 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Finding the Eccentricit of a Hperbola Eccentricit is a ratio that indicates the appearance of a conic. DEFINITION The eccentricit e of a hperbola is the ratio e = c a. PROPERTIES OF ECCENTRICITY FOR HYPERBOLAS For all hperbolas, e > 1 because c > a. If c is much greater than a, then e is large, and the branches of the hperbola are wide. If c is close to a, then e is close to 1, and the branches of the hperbola are narrow. TIP Ellipses and hperbolas have the same ratio for eccentricit. e = a c center a F 1 V center V 1 F c V F1 a c V 1 F The values of c and e are large. The hperbola has wide branches. The values of c and e are close to 1. The hperbola has narrow branches. Eample For the equation ( ) ( + ), do the following: 1 1 A Find the coordinates of the foci. B Find the eccentricit. Solution The equation is in graphing form ( k) h) ( a b and indicates that the hperbola has a vertical transverse ais. A To find the coordinates of the foci, find c. a = b, so a = b =. To find c, substitute these values into the equation: c = a + b + 1 = 3. So c = 3 5.. The coordinates of the foci are approimatel (, 7.) and (, 3.). B The eccentricit for this hperbola is e = a c = 3 1.. 8 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Problem Set Find the equation of the hperbola. 1.. 8 (0, ) ( 5, 0) (5, 0) 8 8 (0, ) 8 1 1 8 (5, 11) (3, ) (7, ) 3 8 1 (5, 3). 5. 1 8 1 (, 8) 8 (0, 5) (, ) (, 5) 8 (, ) ( 11, ) ( 1, ) (, 0) 1 1 1 8 3.. 8 (5, ) 8 8 1 1 1 (, ) (1, ) 8 1 (5, ) 1 1 18 9 ( 7, 3) 9 ( 1, ) 3 3 3 (5, 3) ( 1, 8) 1 15 HYPERBOLAS 9 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
For each problem, do the following: A. Find the coordinates of the center. B. Find the length of the transverse ais. C. Find the length of the conjugate ais. D. Find the equations of the asmptotes. E. Graph the hperbola. 7. 8. 9.. 11. 1. 0 5 9 81 1 ( 11) 9 ( ) 9 ( + 3) ( 5) 1 81 13. 1. 15. 1. 17. 18. _ ( 1) _ 3 ( ) 9 ( + ) 11 ( 1) 0 ( ) _ ( + ) 81 9 ( 1) ( + 1) 5 1 ( + 9) 9 ( + 5) 85 ( 1) ( ) 3 Find the coordinates of the foci of the hperbola. 19. 0. ( 1) 3 ( + 7) 1 9 1.. ( + 8) ( + ) 9 _ ( 1) ( 3) 1 Find the eccentricit of the hperbola. 3. 3 5. ( 3) ( 8) 5.. 8 (0, 3) ( 3, 0) (3, 0) 8 8 (0, 3) 8 8 (1, ) (, ) (, ) 8 (7, ) 8 1 30 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Answer each question. 7. The paths of two comets are represented b the hperbola with equation 1 9. Each branch represents the path of one comet. Find one point along each path at which the comets paths are closest. 8. The shapes of a pair of hperbolic mirrors can be described b the graph of _ ( ) ( 3) 9. An object is placed in the center between the mirrors. Find the location of the object. 9. If a part of a satellite s path can be modeled b the graph of a hperbola, find the equation of the hperbola, given a = 0,531 km and c = 55,31 km. Assume the center of the hperbola is at the origin and the hperbola has a horizontal transverse ais. 30. During storm conditions, an airplane is forced to travel along the path modeled b the right branch of the hperbola with equation 1 5. Find the originall planned path if it corresponds to the positivel sloped asmptote. * 31. Challenge A meteor travels along the path of a hperbola with the equation ( ) 900 500. Find each of the following: A. the lengths of the transverse and conjugate aes B. the location of the center of the hperbola C. the coordinates of the foci D. the slopes of the asmptotes * 3. Challenge Using the diagram, show that d d 1 = a. Hint: Choose a verte of the hperbola for the point (, ). 8 (, ) d 1 focus d 8 8 focus 8 HYPERBOLAS 31 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Parabolas You have learned that the graph of a quadratic function is a parabola that opens upward or downward. Now ou ll learn more about parabolas, including those that open left and right (and therefore do not represent functions). Deriving an Equation of a Parabola from the Definition DEFINITIONS A parabola is the set of all points in a plane that are equidistant from a fied line (the directri) and a fied point (the focus). focus d 1 (, ) The standard equation of a parabola can be derived b using the definition. The parabola below has focus (0, 3) and directri = 3. The distance from a point to a line is the length of the line segment from the point perpendicular to the line. So the distance from (, ) to the horizontal directri is the vertical distance from (, ) to (, 3). directri d d 1 = d for an point (, ) on the parabola. focus (0, 3) d 1 (, ) d 1 = d ( 0) + ( 3) = ( ) + ( + 3) + + 9 = + + 9 + + 9 = + + 9 Distance Formula Epand binomials. Square each side. Simplif. d directri = 3 (, 3) 3 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
} } Writing an Equation of a Parabola, Given Its Graph There are man equivalent was to write the equation. You could write it in the general form of a second-degree equation: 1 = 0. Or, because the parabola represents a function, ou could also write it in the standard form of a quadratic function: = 1 1. Ever parabola has a single verte that lies halfwa between the focus and directri. The ais of smmetr is the line that passes through the focus and verte, perpendicular to the directri. The ais of smmetr separates the graph into two halves that are reflection images of each other. In the diagrams below, V is the verte with coordinates (h, k), F is the focus, and f is the directed distance from V to F (positive for right or up; negative for left or down). directri focus verte (, ) ais of smmetr Parabola with horizontal ais of smmetr: Parabola with vertical ais of smmetr: directri (, ) (, ) directri directri ais of smmetr ais of smmetr V(h, k) f > 0 F f < 0 F V(h, k) ais of smmetr F f > } 0 (, ) V(h, k) directri ais of smmetr V(h, k) F (, ) f < 0 } Equation in Graphing Form: h = a( k) Equation in Graphing Form: k = a( h) PROPERTIES OF PARABOLAS If the directed distance from verte to focus (the focal distance) is represented b f, then the directed distance from verte to directri is f, and a f ( or f = a 1 ). f 0 and a 0. f > 0 and a > 0 if the parabola opens right or up. f < 0 and a < 0 if the parabola opens left or down. The eccentricit of a parabola is e. TIP The eccentricit of a parabola is e = XF, where X is an point on XD the parabola, F is the focus, and D is the directri. Because XF = XD for an point X on a parabola, e. PARABOLAS 33 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Eample 1 Write the equation in graphing form of the parabola shown. Solution The ais of smmetr has the equation = 1, and it intersects the directri at (0.5, 1). The verte (h, k) is halfwa between the focus and directri, so it is the midpoint of the segment from (0.375, 1) to (0.5, 1): 0.375 + 0.5 (h, k) = ( 1 + ( 1), ) = ( 1, ) = (0.5, 1). The focus is 0.15 unit left of the verte, so f = 0.15. Find a. Substitute h, k, and a into the equation in graphing form for a horizontal parabola. a f = 1 ( 0.15) = h = a( k) 0.5 = ( + 1) 1 1 focus (0.375, 1) 1 (0.5, 1) directri = 0.5 (0.5, 1) = 1 ais of smmetr Graphing a Parabola, Given Its Equation in Graphing Form Eample Graph the parabola with equation 3 = ( ). Solution Use the same techniques that ou learned to graph the verte form of a quadratic function. The equation is in the graphing form h = a( k), so the parabola has a horizontal ais of smmetr. Since the value of a is positive, the parabola opens to the right. Step 1 Plot the verte (3, ) and sketch the ais of smmetr =. Step Find and plot two more solutions to the equation 3 = ( ), such as (5, 5) and (8, 5.58). Step 3 Reflect the points ou found in Step across the ais of smmetr to find two points on the other side: (5, 3) and (3, 3.58). Step Draw a smooth curve through all the points. THINK ABOUT IT In the stud of conic sections, h = a( k) and k = a( h) are called equations in graphing form of parabolas. Equations of the form k = a( h) represent functions and can be rewritten in the form = a + b + c, which is called standard form in the stud of quadratic functions. (Tr it. You should find that b = ah and c = ah + k.) 5 (3, ) (5, 5) (5, 3) (8, 5.58) = (8, 3.58) TIP The focus, directri, and ais of smmetr are not reall part of the graph, but the provide important information. If ou identif them in a problem, it is useful to include them in the graph. 8 3 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Identifing the Verte, Focal Length, Focus, and Directri of a Parabola, Given Its Equation Eample 3 The equation of a parabola is = 0.15( + 3). A Identif the coordinates of the verte. B Identif the focal length and the coordinates of the focus. C Identif the equation of the directri. D Graph the parabola. Label the verte, focus, ais of smmetr, directri, and two other points on the curve. Solution The value of a is positive, so the parabola opens upward. A The equation is in the graphing form k = a( h), with h = 3, k =, and a = 0.15. So the verte (h, k) is at ( 3, ). B The distance from the verte to the focus is f = a 1 = _ 1 0.15 = 0.5 1 =. Because f =, the focus is units above the verte at F( 3, ). C Because f =, the directri is units below the verte and passes through ( 3, ). The ais of smmetr is vertical, so the directri is horizontal. The equation of the directri is =. D = 0.15 ( + 3) ( 7, ) = = 3 F ( 3, ) V( 3, ) 1 8 8 (1, ) PARABOLAS 35 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Problem Set Find the equation of the parabola. 1. 1. 1 1 1 3 5 F( 3.15, ) V( 3, ) 3 5 V(1, ) ( ) 1 F 1,. 5. 1 5 3 1 1 1 V( 3, ) 8 8 F 3, 1 ( 8 ) 3 5 F( 1,.75) 8 = 5.5 3. 8..75 F(, 5 ) 1 1 3 5 7 8 9 F(7, 3) V (7, ) 8 3 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
For each problem, do the following: A. Identif the coordinates of the verte. B. Graph the parabola. 7. = 8. = ( + 3) 9. + 1. = ( 7). + = 5( ) 11. 5 = ( 3) 1. = ( 1) 13. 1 = 1. + = 1 ( ) 15. + = ( + 1) 1. + 1.5 = 5( 7) For each problem, do the following: A. Identif the focal length. B. Identif the coordinates of the focus. C. Find the equation of the directri. 17. 1 = 3( + 1) 18. = (.5) 19. 5 ( 9) 0. 1 = 1 ( 8) Solve. 5. The flame of the Olmpic torch is sparked using a parabolic mirror. If the shape of the parabolic mirror is modeled b the equation + = 0.5( 3), find the coordinates of the focus.. A parabolic reflector in a telescope is modeled b the equation = 0.5( + 1). Graph the parabola and label its focus point. 7. Olivia hits a game-winning home run. The ball travels in the parabolic path shown in the graph. Write an equation for the path of the baseball in the air. 1. + 0 = 0.( ). = 0.75( + 5) 3. + 7 = 5( 1). + = 1 ( 3) 8 8. Andrew launches a to rocket into the air and watches it travel along a parabolic path modeled b the equation 5 = 5( ). Find the verte, focus, and directri of the parabola. * 9. Challenge The focus of a parabola is ( 1, ) and the directri is = 1. Write the equation of the parabola in graphing form. 30. Jerem is building a bird feeder like the one shown. The wire he is using to hang the bird feeder passes through the focus of each end. Write the equation representing the shape of each end of the bird feeder. (3, ) 1 1 3 5 7 8 9 11 Verte (0, 0) Focus (0, ) PARABOLAS 37 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Putting Conics into Graphing Form Conic sections have man applications in science and technolog. PROPERTIES OF A + B + C + D + E + F = 0 The general form of the equation for an conic section is A + B + C + D + E + F = 0 where,, A, B, C, D, E, and F are all real numbers. Use the value of the discriminant B AC to determine the tpe of conic described b the general form. Discriminant Value B AC > 0 B AC = 0 B AC < 0 Conic Section Hperbola Parabola Ellipse if A C Circle if A = C THINK ABOUT IT This definition of the value of the discriminant is different than the one used for determining the tpe of roots of a parabola. Identifing a Conic from Its Equation in General Form Eample 1 equation. Use the discriminant to identif the conic section in each A + + 1 35 = 0 B 5 + 0 1 + 98 = 0 Solution Solution A =, B = 0, C = A = 5, B = 0, C = B AC = 0 B AC = 0 5 = 0 = < 0 = 0 0 = 0 < 0 A = C = A = 5 = C The graph is a circle. The graph is an ellipse. C 9 18 9 = 0 D 3 1 + 13 = 0 Solution Solution A =, B = 0, C = 9 A = 0, B = 0, C = 3 B AC = 0 ( 9) B AC = 0 0 3 = 0 + 1 > 0 = 0 0 = 0 The graph is a hperbola. The graph is a parabola. 38 UNIT 15 CONIC SECTIONS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Converting an Equation of a Circle from General Form to Graphing Form When the general second-degree equation A + B + C + D + E + F = 0 has B AC < 0 and A = C (but not zero), the equation is that of a circle. Converting from general form to graphing form will help ou find the center and radius of the circle. Eample An equation of a circle is + + 1 1 = 0. A Write the equation of the circle in graphing form. B Graph the circle. Solution The equation is in general second-degree form with B AC < 0 and A = C, so this is an equation of a circle. A To write the equation in graphing form, complete the square for each variable. + + 1 1 = 0 Equation in general form + + 1 Isolate the constant. + + 1 Group like terms. ( + ) + ( ) Factor to get leading terms and. ( + + 9) + ( + ) + 3 + 1 Complete the squares. Because ou added 9 and inside the parentheses, ou are reall adding 9 = 3 and. ( + 3) + ( ) = Write the trinomials as squared binomials. ( + 3) + ( ) Divide ea ch side b. B In graphing form, h = 3, k =, and r =. Step 1 First plot the center (h, k) at ( 3, ). Step Count units left, right, up, and down from the center to find the approimate locations of four points on the circle. Step 3 Draw the circle through the four points. 8 ( 3, ) ( 7, ) C( 3, ) (1, ) 8 ( 3, ) PUTTING CONICS INTO GRAPHING FORM 39 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.