TASK 7.3.4: CONICS IN GENERAL FORM AX 2 + CY 2 + DX + EY + F = 0

Transcription

1 1 TASK 7.3.4: CONICS IN GENERAL FORM AX 2 + CY 2 + DX + EY + F = 0 Solutions Graph the following conics using the program CONICSTU. What type of conic is formed when the equation is graphed? Give the coordinates of the center of the conic (the vertex on parabolas). Type of Conic Center 1. x 2 + y 2 25 = 0 Circle ( 0, 0 )_ 2. x 2 y 2 25 = 0 Hyperbola ( 0, 0 )_ 3. 4x 2 9y 2 36 = 0 _Hyperbola ( 0, 0 )_ 4. 4x 2 + 9y 2 36 = 0 Ellipse ( 0, 0 )_ 5. y 2 4x =0 Parabola ( 0, 0 )_ 6. x 2 + y 2 4x 2y 19 = 0 Circle ( 2, 1 )_ 7. x 2 y 2 + 2x +6y 33 = 0 Hyperbola ( -1, 3 )_ 8. y 2 4x 6y +13 =0 _Parabola ( 1, 3 )_ 9. 4x 2 + 9y 2 24x +18y + 9 = 0 Ellipse ( 3, -1 )_ 10. x 2 + y 2 + 6x 2y 8 = 0 Circle ( -3, 1 )_ 1. Circle Center ( 0,0) 2. Hyperbola Center ( 0,0)

2 2 3. Hyperbola Center ( 0,0) 4. Ellipse Center ( 0,0) 5. Parabola Vertex ( 0,0) 6. Circle Center ( 2,1) 7. Hyperbola Center ( -1,3) 8. Parabola Vertex ( 1,3) 9. Ellipse Center ( 3,-1) 10. Circle Center ( -3,1) Which constants affect the type of conic that is graphed? The constants that seem to affect the graph most are A and C. D, E, F affect the elongation of an ellipse and whether parabolas and hyperbolas open up, down, left, right.

3 3 What conclusions can you reach? If A = C, then a circle is graphed. If A C, A>0 and C>0, then an ellipse is graphed. If A C, A<0 and C>0, then a hyperbola is graphed. If A C, A>0 and C<0, then a hyperbola is graphed. If A or C is zero and the other is not, then a parabola is graphed. **Remember that these conclusions do not work for the general conic containing a Bxy term. Use the method of "completing the square" to write the standard form equations for #6-#10 on the previous page. Your graphs may help you with this. 6. x 2 + y 2 4x 2y 19 = 0 7. x 2 y 2 + 2x +6y 33 = 0 (x 2) 2 + (y 1) 2 = 25 ( x + 1) 2 25 (! y! 3 ) 2 25 = 1 8. y 2 4x 6y +13 =0 9. 4x 2 + 9y 2 24x +18y + 9 = 0 (y-3) 2 = 4(x-1) ( x! 3) 2 9 ( + y + 1 ) 2 4 = x 2 + y 2 + 6x 2y 8 = 0 (x +3) 2 + (y 1) 2 = 18 Students may look at the graph and think that the radius is "4". After working this problem, they can trace on the graph to confirm that the radius was not "4" but rather 18. Math notes Remember that the conclusions reached in this task do not work for the general conic containing a Bxy term. However, they are important for students to know when changing to standard form by completing the square. Some discussion could occur with students concerning B 2-4AC at this time. Students will have probably already asked where the B term is?

4 4 Teaching notes This task could be used as an assessment. Students will need the program CONICSTU. It should be available for download from the CD. A copy of the program is provided at the end of these notes. If this task is not offered as an assessment, participants should work in groups to complete the first page of the task. Note: The window is set for them in the program because a "square window" is essential to visually distinguish between circles and ellipses. Students may change the window with a zoom, but it will revert to the programmed window with the next graph. Participants may want to plot their guess for center using the STAT PLOT. A point can be entered and the plot turned on after the program initially draws the graph. Then push the GRAPH key. Discuss how they can recognize the type of conic graphed just by looking at the equation. To introduce the second page of the activity, relate the graph of #6 to its standard form equation. Work "backward" from the standard form to the general form given on the task. Relate the (x - 2) 2 term of the standard form to x 2 4x in the general form. Discuss the process "completing the square" and how knowing the center might help with this or confirm the answer once it is obtained. Work #6 and #9 together, then assign the other problems to the students. Allow them to work together and expect many questions. Closely monitor their progress and be prepared to help struggling students. Encourage participants to confirm their work with the characteristics of the graphs from page 1. Look for centers, radii, lengths of major and minor axes, etc. Technology notes Due to character transmission issues, the CONICSTU program might not run properly if it is copied and pasted (and then sent to your TI device) electronically. If the program does not run, there are two things that need to be fixed on the calculator. Edit the program and make these changes: 1. Insert negative signs where appropriate. In transmission, negative signs were changed to subtraction signs 2. Insert function variables where appropriate (Y1, Y2, Y3, Y4). In transmission, these variables were separated into two distinct characters. Go to VARS then Y-VARS then FUNCTION and insert the appropriate variable

5 5 TI Program: CONICSTU PlotsOff FnOff AxesOff Text(8,0,"TO GRAPH THE ") Text(18,0,"GENERAL CONIC") Text(28,0,"AX^2+CY^2+DX+EY+F=0") Text(38,0,"ENTER THE COEFFICIENTS") Text(55,68,"[ENTER]") Pause FnOff PlotsOff AxesOn Simul Func -15 Xmin 15 Xmax 1 Xscl -10 Ymin 10 Ymax 1 Yscl Connected Simul Disp "ENTER" Disp "COEFFICIENTS" Disp "A" Input A Disp "C" Input C Disp "D" Input D Disp "E" Input E Disp "F" Input F If C=0 Then "(-AX^2-DX-F)/E" Y 4 :End " ((E)^2-4C(AX^2+DX+F))" Y 1 "(-(E)+Y 1 )/(2C)" Y 2 "(-(E)-Y 1 )/(2C) Y 3 FnOff 1 DispGraph

6 6 TASK 7.3.4: CONICS IN GENERAL FORM AX 2 + CY 2 + DX + EY + F = 0 Graph the following conics using the program CONICSTU. What type of conic is formed when the equation is graphed? Give the coordinates of the center of the conic (the vertex on parabolas). Type of Conic Center 1. x 2 + y 2 25 = 0 2. x 2 y 2 25 = x 2 9y 2 36 = x 2 + 9y 2 36 = 0 5. y 2 4x =0 6. x 2 + y 2 4x 2y 19 = 0 7. x 2 y 2 + 2x +6y 33 = 0 8. y 2 4x 6y +13 =0 9. 4x 2 + 9y 2 24x +18y + 9 = x 2 + y 2 + 6x 2y 8 = 0 Which constants affect the type of conic that is graphed? What conclusions can you reach?

7 7 Use the method of "completing the square" to write the standard form equations for #6-#10 on the previous page. Your graphs may help you with this. 6. x 2 + y 2 4x 2y 19 = 0 7. x 2 y 2 + 2x +6y 33 = 0 8. y 2 4x 6y +13 =0 9. 4x 2 + 9y 2 24x +18y + 9 = x 2 + y 2 + 6x 2y 8 = 0