11X1 T12 07 chord of contact (2011)

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<ol><li> 1. Chord of Contact </li><li> 2. Chord of Contact y x 2 4ay x </li><li> 3. Chord of Contact y x 2 4ay We know the coordinates of an external point (T) x </li><li> 4. Chord of Contacty x 2 4ayWe know the coordinatesof an external point (T)xT x0 , y0 </li><li> 5. Chord of Contacty x 2 4ayWe know the coordinatesof an external point (T)From this external point,x two tangents can be drawnmeeting the parabola at PT x0 , y0 and Q. </li><li> 6. Chord of Contacty x 2 4ayP(2ap, ap 2 )We know the coordinatesof an external point (T)Q(2aq, aq 2 ) From this external point,x two tangents can be drawnmeeting the parabola at PT x0 , y0 and Q. </li><li> 7. Chord of Contacty x 2 4ayP(2ap, ap 2 )We know the coordinatesof an external point (T)Q(2aq, aq 2 ) From this external point,x two tangents can be drawnmeeting the parabola at PT x0 , y0 and Q. </li><li> 8. Chord of Contacty x 2 4ayP(2ap, ap 2 )We know the coordinatesof an external point (T)Q(2aq, aq 2 ) From this external point,x two tangents can be drawnmeeting the parabola at PT x0 , y0 and Q.The line joining these twopoints is called the chordof contact. </li><li> 9. Chord of Contacty x 2 4ayP(2ap, ap 2 )We know the coordinatesof an external point (T)Q(2aq, aq 2 ) From this external point,x two tangents can be drawnmeeting the parabola at PT x0 , y0 and Q.The line joining these twopoints is called the chordof contact. </li><li> 10. (1) Parametric approach </li><li> 11. (1) Parametric approach1 Show that PQ has equation p q x 2 y 2apq </li><li> 12. (1) Parametric approach1 Show that PQ has equation p q x 2 y 2apq2 Show the two tangents have equationspx y ap 2 0 and qx y aq 2 0 </li><li> 13. (1) Parametric approach1 Show that PQ has equation p q x 2 y 2apq2 Show the two tangents have equationspx y ap 2 0 and qx y aq 2 0 3 Show that T is the point a p q , apq </li><li> 14. (1) Parametric approach1 Show that PQ has equation p q x 2 y 2apq2 Show the two tangents have equationspx y ap 2 0 and qx y aq 2 0 3 Show that T is the point a p q , apq 4 But T is x0 , y0 </li><li> 15. (1) Parametric approach1 Show that PQ has equation p q x 2 y 2apq2 Show the two tangents have equationspx y ap 2 0 and qx y aq 2 0 3 Show that T is the point a p q , apq 4 But T is x0 , y0 x0 a p q x0pq a </li><li> 16. (1) Parametric approach1 Show that PQ has equation p q x 2 y 2apq2 Show the two tangents have equationspx y ap 2 0 and qx y aq 2 0 3 Show that T is the point a p q , apq 4 But T is x0 , y0 x0 a p q y0 apqx0pq a </li><li> 17. (1) Parametric approach1 Show that PQ has equation p q x 2 y 2apq2 Show the two tangents have equationspx y ap 2 0 and qx y aq 2 0 3 Show that T is the point a p q , apq 4 But T is x0 , y0 x0 a p q y0 apqx0pq a x0 x PQ is 2 y 2 y0a </li><li> 18. (1) Parametric approach1 Show that PQ has equation p q x 2 y 2apq2 Show the two tangents have equationspx y ap 2 0 and qx y aq 2 0 3 Show that T is the point a p q , apq 4 But T is x0 , y0 x0 a p q y0 apqx0pq a x0 x PQ is 2 y 2 y0aHence the chord of contact is x0 x 2a y0 y </li><li> 19. (1) Parametric approach1 Show that PQ has equation p q x 2 y 2apq2 Show the two tangents have equationspx y ap 2 0 and qx y aq 2 0 3 Show that T is the point a p q , apq 4 But T is x0 , y0 x0 a p q y0 apqx0pq a x0 x notice PQ is 2 y 2 y0a similarityto tangentHence the chord of contact is x0 x 2a y0 y </li><li> 20. (2) Cartesian approach </li><li> 21. (2) Cartesian approach 1 Show that PT has equation xx1 2a y y1 </li><li> 22. (2) Cartesian approach 1 Show that PT has equation xx1 2a y y1 T lies on PT x0 x1 2a y0 y1 </li><li> 23. (2) Cartesian approach 1Show that PT has equation xx1 2a y y1 T lies on PT x0 x1 2a y0 y1 P x1 , y1 lies on the line with equation x0 x 2a y0 y </li><li> 24. (2) Cartesian approach 1Show that PT has equation xx1 2a y y1 T lies on PT x0 x1 2a y0 y1 P x1 , y1 lies on the line with equation x0 x 2a y0 y 2 Show that QT has equation xx2 2a y y2 </li><li> 25. (2) Cartesian approach 1Show that PT has equation xx1 2a y y1 T lies on PT x0 x1 2a y0 y1 P x1 , y1 lies on the line with equation x0 x 2a y0 y 2 Show that QT has equation xx2 2a y y2 T lies on QT x0 x2 2a y0 y2 </li><li> 26. (2) Cartesian approach 1Show that PT has equation xx1 2a y y1 T lies on PT x0 x1 2a y0 y1 P x1 , y1 lies on the line with equation x0 x 2a y0 y 2 Show that QT has equation xx2 2a y y2 T lies on QT x0 x2 2a y0 y2 Q x2 , y2 lies on the line with equation x0 x 2a y0 y </li><li> 27. (2) Cartesian approach 1Show that PT has equation xx1 2a y y1 T lies on PT x0 x1 2a y0 y1 P x1 , y1 lies on the line with equation x0 x 2a y0 y 2 Show that QT has equation xx2 2a y y2 T lies on QT x0 x2 2a y0 y2 Q x2 , y2 lies on the line with equation x0 x 2a y0 y Hence the chord of contact is x0 x 2a y0 y </li><li> 28. (2) Cartesian approach 1Show that PT has equation xx1 2a y y1 T lies on PT x0 x1 2a y0 y1 P x1 , y1 lies on the line with equation x0 x 2a y0 y 2 Show that QT has equation xx2 2a y y2 T lies on QT x0 x2 2a y0 y2 Q x2 , y2 lies on the line with equation x0 x 2a y0 y Hence the chord of contact is x0 x 2a y0 y Exercise 9H; 1c, 2d, 3, 6, 8, 10, 14</li></ol>