8.9 Intersection of Lines and Conics



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8.9 Intersection of Lines and Conics The centre circle of a hockey rink has a radius of 4.5 m. A diameter of the centre circle lies on the centre red line. centre (red) line centre circle INVESTIGATE & INQUIRE 1. A coordinate grid is superimposed on a hockey rink so that the origin is at the centre of the centre circle and the x-axis lies along the red line. The side length of each grid square represents 1 m. a) Write the equation of the centre circle. b) Graph the centre circle on grid paper or use a graphing calculator. 2. The puck is passed across the centre line along each of the following lines. Graph each line and find the number of points in which it intersects the centre circle. Find the coordinates of any points of intersection, rounding to the nearest tenth, if necessary. a) y x b) y 8 x c) y 4.5 3. a) What is the maximum number of points in which a line and a circle can intersect? b) What is the minimum number of points in which a line and a circle can intersect? 4. Without graphing each of the following lines, state the number of points in which it intersects the centre circle. a) y 6 b) y 4.5 c) y 1 d) y 3x 8.9 Intersection of Lines and Conics MHR 675

Recall that a system of equations consists of two or more equations that are considered together. If the graphs of two equations in a system are a straight line and a conic, the system will have no solution, one solution, or two solutions. y y y 0 x 0 x 0 x one solution two solutions no solution EXAMPLE 1 Finding Points of Intersection of a Line and a Circle Find the coordinates of the points of intersection of the line y x 1 and the circle x 2 + y 2 25. SOLUTION 1 Paper-and-Pencil Method Solve the system of equations by substitution. y x 1 (1) x 2 + y 2 25 (2) Substitute x 1 for y in (2): x 2 + (x 1) 2 25 Expand: x 2 + x 2 2x + 1 25 Simplify: 2x 2 2x 24 0 Divide both sides by 2: x 2 x 12 0 Factor: (x 4)(x + 3) 0 Use the zero product property: x 4 0 or x + 3 0 x 4 or x 3 Substitute these values of x into (1). For x 4, For x 3, y x 1 y x 1 4 1 3 1 3 4 676 MHR Chapter 8

Check in (1). Check in (2). For (4, 3), For (4, 3), L.S. y R.S. x 1 L.S. x 2 + y 2 R.S. 25 3 4 1 4 2 + 3 2 3 25 For ( 3, 4) For ( 3, 4), L.S. y R.S. x 1 L.S. x 2 + y 2 R.S. 25 4 3 1 ( 3) 2 +( 4) 2 4 25 The coordinates of the points of intersection are (4, 3) and ( 3, 4). SOLUTION 2 Graphing-Calculator Method Solve the equation of the circle for y. x 2 + y 2 25 y 2 25 x 2 y ± 25 x 2 So, in the same viewing window, graph y x 1, y 25, x 2 and y 25. x 2 To make the circle look like a circle, use the Zsquare instruction. To find the points of intersection, use the intersect operation. The coordinates of the points of intersection are ( 3, 4) and (4, 3). These coordinates can be checked by substitution into the original equations. EXAMPLE 2 Finding Points of Intersection of a Line and an Ellipse Find the coordinates of the points of intersection of the line 2x + y 1 and the ellipse 4x 2 + y 2 25. 8.9 Intersection of Lines and Conics MHR 677

SOLUTION Solve the system of equations by substitution. 2x + y 1 (1) 4x 2 + y 2 25 (2) Solve for y in (1): 2x + y 1 y 1 2x Substitute 1 2x for y in (2): 4x 2 + (1 2x) 2 25 Expand: 4x 2 + 1 4x + 4x 2 25 Simplify: 8x 2 4x 24 0 Divide both sides by 4: 2x 2 x 6 0 Factor: (2x + 3)(x 2) 0 Use the zero product property: 2x + 3 0 or x 2 0 Substitute for x in (1) to find y. For x 3 2, For x 2, 2 3 2(2) + y 1 2 + y 1 4 + y 1 y 1 4 3 + y 1 3 y 4 Check in (1). x 3 2 or x 2 For 3 2, 4, For (2, 3), L.S. 2x + y R.S. 1 L.S. 2x + y R.S. 1 2 3 2 + 4 2(2) 3 1 Check in (2). For 3 2, 4, 1 For (2, 3), L.S. 4x 2 + y 2 R.S. 25 L.S. 4x 2 + y 2 R.S. 25 4 3 2 2 + 4 2 4(2) 2 + ( 3) 2 The system can be modelled graphically. 16 + 9 9 + 16 25 25 The coordinates of the points of intersection are 3 2, 4 and (2, 3). 678 MHR Chapter 8

EXAMPLE 3 Finding Points of Intersection of a Line and a Parabola Find the coordinates of the points of intersection of the parabola y 4 (x + 1) 2 and a) the line y 4x + 4 b) the line y 3x + 13 SOLUTION a) Solve the system of equations by substitution. y 4x + 4 (1) y 4 (x + 1) 2 (2) Substitute 4x + 4 for y in (2): 4x + 4 4 (x + 1) 2 Expand: 4x + 4 4 (x 2 + 2x + 1) Simplify: 4x x 2 2x 1 x 2 2x + 1 0 Factor: (x 1)(x 1) 0 Use the zero product property: x 1 0orx 1 0 x 1 or x 1 Substitute for x in (1) to find y. y 4x + 4 4(1) + 4 0 Check in (1). L.S. y R.S. 4x + 4 0 4(1) + 4 0 Check in (2). L.S. y 4 R.S. (x + 1) 2 0 4 (1 + 1) 2 4 4 The coordinates of the point of intersection are (1, 0). The system can be modelled graphically. 8.9 Intersection of Lines and Conics MHR 679

b) Solve the system of equations. y 3x + 13 (1) y 4 (x + 1) 2 (2) Substitute 3x + 13 for y in (2): 3x + 13 4 (x + 1) 2 Expand: 3x + 13 4 (x 2 + 2x + 1) Simplify: 3x + 9 x 2 2x 1 x 2 + 5x + 10 0 For x 2 + 5x + 10 0, a 1, b 5, and c 10. Substitute these values into the quadratic formula. x b ± b c 2 4a 2a 5 ± 5 1)(10) 2 4( 2(1) 5 ± 25 0 4 2 5 ± 15 2 Since no real number is the square root of a negative number, there are no real solutions. The parabola and the line do not intersect. The system can be modelled graphically. EXAMPLE 4 Finding Points of Intersection of a Line and a Hyperbola Find the coordinates of the points of intersection of the line y x + 3 and the hyperbola y 2 4x 2 4. Round answers to the nearest tenth. SOLUTION Solve the system of equations by substitution. y x + 3 (1) y 2 4x 2 4 (2) Substitute x + 3 for y in (2): (x + 3) 2 4x 2 4 Expand: x 2 + 6x + 9 4x 2 4 Simplify: 3x 2 + 6x + 5 0 Multiply by 1: 3x 2 6x 5 0 680 MHR Chapter 8

Solve using the quadratic formula. x b ± b c 2 4a 2a 6 ± ( 6) 2 5) 4(3)( 2(3) 6 ± 36 0 + 6 6 6 ± 96 6 2.6 or 0.6 The system can be modelled graphically. Substitute for x in (1) to find y. For x 2.6, y 2.6 + 3 For x 0.6,y 0.6 + 3 5.6 2.4 The coordinates of the points of intersection are (2.6, 5.6) and ( 0.6, 2.4), to the nearest tenth. EXAMPLE 5 Flight Path A pilot is flying a small aircraft at 200 km/h directly toward the centre of an intense weather system. She decides to change direction to avoid the worst of the weather. Taking the aircraft as the origin of a coordinate grid, with the side length of each grid square representing 1 km, the weather system can be modelled by the relation (x 20) 2 + (y 10) 2 49. The new direction of the aircraft is along the line y 0.2x in the first quadrant. Will the aircraft completely avoid the weather system? If not, for what amount of time will the aircraft be within the weather system, to the nearest tenth of a minute? SOLUTION 1 Paper-and-Pencil Method Solve the system of equations. (x 20) 2 + (y 10) 2 49 (1) y 0.2x (2) Substitute 0.2x for y in (1). (x 20) 2 + (0.2x 10) 2 49 x 2 40x + 400 + 0.04x 2 4x + 100 49 1.04x 2 44x + 451 0 8.9 Intersection of Lines and Conics MHR 681

Solve using the quadratic formula. x b ± b c 2 4a 2a 44 ± ( 44) 2 4)(451 4(1.0 ) 2(1.04) So, x 24.87 or x 17.43. Substitute in (2). For x 24.87, For x 17.43, y 0.2(24.87) y 0.2(17.43) 4.97 3.49 The coordinates of the points of intersection are approximately (17.43, 3.49) and (24.87, 4.97). So, the aircraft will not completely avoid the weather system. Use the formula for the length of a line segment to find the distance between the two points of intersection. d (x 2 x 1 ) 2 + (y 2 y 1 ) 2 (24.87 3) 17.4.97 2 + (4.49) 3 2 7.59 The aircraft will be within the weather system for a distance of about 7.59 km. Use time distance to find the amount of time within the weather system. speed time 7.59 200 0.037 95 Convert this number of hours to minutes. 0.037 95 60 2.277 So, the aircraft will be within the weather system for 2.3 min, to the nearest tenth of a minute. 682 MHR Chapter 8

SOLUTION 2 Graphing-Calculator Method Solve (x 20) 2 + (y 10) 2 49 for y. (x 20) 2 + (y 10) 2 49 (y 10) 2 49 (x 20) 2 y 10 ± 49 (x 20) 2 y 10 ± 49 (x 20) 2 In the same viewing window, graph y 0.2x, y 10 + 49 (x, 20) 2 and y 10 49 (x. 20) 2 Adjust the window variables and use the Zsquare instruction. To find the points of intersection, use the Intersect operation. The window variables include Xmin 4.8, Xmax 35.2, Ymin 0, Ymax 20. The coordinates of the points of intersection are approximately (17.43, 3.49) and (24.87, 4.97). So, the aircraft will not completely avoid the weather system. The time that the aircraft will be within the weather system can be found using the formula for the length of a line segment and the formula time distance, as shown in Solution 1. speed The aircraft will be within the weather system for 2.3 min, to the nearest tenth of a minute. 8.9 Intersection of Lines and Conics MHR 683

Key Concepts If the graphs of two equations in a system are a straight line and a conic, the system will have no solution, one solution, or two solutions. To find the points of intersection of a linear equation and a quadratic equation algebraically, a) solve the linear equation for one variable b) substitute the expression for that variable in the quadratic equation c) solve the quadratic equation to find any real value(s) of the remaining variable d) substitute any real value(s) of the variable found from part c) into the linear equation to find the value(s) of the other variable To solve a system of a straight line and a conic graphically, graph both equations and determine the coordinates of the point or points of intersection, if they exist. Communicate Your Understanding 1. Describe how you would find the points of intersection of x y 2 and x 2 + y 2 16 algebraically. 2. Describe how you would solve the system 2x 2 + y 2 36 and 2x y 1 graphically. 3. If the graphs of a linear equation and a quadratic equation do not intersect, describe the algebraic solution to the system. Practise A 1. Solve each system of equations. a) y 9 0 b) y + 25 0 y x 2 0 x 2 + y 0 c) y 2x d) y x 0 y x 2 0 x 2 + y 2 32 0 e) y 4x f) y + 3x 0 x 2 y + 4 0 x 2 + y 2 10 2. Solve each system of equations. a) y x 2 + 3 b) x 2 + y 2 36 3x + y 1 2x + y 12 c) x 2 y 2 9 d) x 2 + 4y 2 16 x y 6 5x y 2 e) 4x 2 + y 2 64 f) y 4x 2 1 2 x + y 10 12x y 9 0 g) y x 5 h) 9x 2 + 4y 2 36 x 2 + y 2 4 2x + y 4 i) x + y 4 y 1 2 x 2 3. Solve each system of equations. Round solutions to the nearest hundredth, if necessary. a) x 2 + y 2 25 b) y x 2 + 5 2x + y 5 x + y 2 684 MHR Chapter 8

c) x 2 + 4y 2 100 d) x 2 y 2 64 x y 5 2x + y 14 e) x 2 + y 2 16 f) 2y x 2 0 x 2y 7 2x + y 2 g) x 2 + y 2 17 h) 25x 2 36y 2 900 x + 2y 2 x + 2y 6 i) 9x 2 + y 2 81 j) x 2 + y 2 13 x + 3y 3 2x y 7 k) x 2 64y 2 1 l) 4x 2 y 2 9 0 x + 8y 0 2x y 0 4. Determine the length of the chord AB in each circle, to the nearest hundredth. a) y B y 0.5x A 0 x (x 5) 2 + (y 5) 2 25 b) y B y 0.5x + 2 A 0 x (x + 1) 2 + (y + 1) 2 9 Apply, Solve, Communicate 5. Air-traffic control A radar screen has a range of 50 km. The screen can be modelled on a coordinate grid as a circle centred at the origin with radius 50. a) Write the equation that describes the edge of the radar screen s range. b) A small aircraft flies over the area covered by the radar at 180 km/h, on a path given by y 0.5x 10. Determine the length of time, to the nearest minute, that the plane is within radar range. B 6. Communication In a city park, a sprinkler waters an area with a circumference that can be modelled by the equation x 2 + y 2 20. The sprinkler is located at the origin. A path through the park can be modelled by the equation 2x y 12. Will the sprinkler spray people walking along the path? Explain. 7. Application The radar detector of a lighthouse has a range that can be modelled by the equation x 2 + y 2 5, if the lighthouse is located at the origin of a grid. A boat is travelling on a path defined by the equation x y 2. Will the boat be detected on the radar screen at the lighthouse? Explain. 8.9 Intersection of Lines and Conics MHR 685

8. UV index The UV index from 08:00 to 18:00 Web Connection on one sunny July day in Central Ontario www.school.mcgrawhill.ca/resources/ could be modelled by the quadratic To learn more about the UV index, visit the function y 0.15(x 13) 2 + 7.6, where above web site. Go to Math Resources, y is the UV index and x is the time of day, then to MATHEMATICS 11, to find out where in hours on the 24-hour clock. to go next. Write a brief report about how a) Write a system of equations to determine the the UV index is determined. time period when the UV index was 7 or more. b) Solve the system of equations from part a) and state the time period. 9. Radar The radar on a marine police launch has a range of 28 km. While the launch is at anchor in a bay, the radar shows a boat travelling on a linear path given by y 0.7x + 20. If the boat is visible on the radar screen for 2 h, at what speed is it travelling, to the nearest kilometre per hour? 10. Power consumption The power consumption of a college cafeteria varies through the day. The power consumption can be approximated by the quadratic function y 0.25(x 12) 2 + 50, where y is the power consumption, in kilowatts (kw), and x is the time of day, in hours on the 24-hour clock. a) Write a system of equations that could be used to determine the times of day when the power consumption is at least 40 kw. b) Solve the system of equations from part a) to find the time period, to the nearest hour. 11. Inquiry/Problem Solving The line x + 3y 5 0 intersects the circle (x 5) 2 + (y 5) 2 25 in two points, A and B. a) Find the coordinates of the endpoints of the chord AB. b) Verify that the right bisector of the chord AB passes through the centre of the circle. 12. In the figure, the circle has its centre at the origin and a radius of 5 units. Determine the exact length of the chord MN. y 5 M y 2x N y 0.5x 5 0 5 x 5 x 2 + y 2 25 686 MHR Chapter 8

13. Theatre A spotlight illuminates an elliptical area on a theatre stage. The shape of the ellipse can be modelled by the equation 2x 2 + y 2 9 on a grid with 1 unit equal to 1 m. An actor is crossing the stage along a path with the equation y 2x 1. Find the length of the actor s path illuminated by the spotlight, to the nearest tenth of a metre. C 14. Hyperbolic mirror A hyperbolic mirror has the shape of one branch of a hyperbola. Light rays directed at the focus of a hyperbolic mirror are reflected toward the other focus of the hyperbola, as shown in the diagram. y F 1 0 F 2 x A mirror can be modelled by the left branch of the hyperbola with the x 2 y 2 equation 1. A light source is located at (0, 8). Find the 9 16 coordinates of the point on the mirror at which light from the source is reflected to the point (5, 0). Round the coordinates to the nearest tenth. 15. Solve each system of equations. a) x + y 2 2 2y 2 2 x( 2 + 2) b) x 2 + y 2 1 y 3x + 1 x 2 + (y + 1) 2 4 8.9 Intersection of Lines and Conics MHR 687