This document discusses finding the equation of the chord of contact of a parabola given an external point. It provides two approaches: (1) using parametric equations to derive the chord of contact equation as x0x = 2a(y0 + y) and (2) using Cartesian coordinates to show that the external point lies on the tangent lines, also deriving the chord of contact equation as x0x = 2a(y0 + y). It then lists related exercises.

1. Chord of Contact

2. Chord of Contact
y x 2  4ay
x

3. Chord of Contact
y x 2  4ay
We know the coordinates
of an external point (T)
x

4. Chord of Contact
y x 2  4ay
We know the coordinates
of an external point (T)
x
T  x0 , y0 

5. Chord of Contact
y x 2  4ay
We know the coordinates
of an external point (T)
From this external point,
x two tangents can be drawn
meeting the parabola at P
T  x0 , y0  and Q.

6. Chord of Contact
y x 2  4ay
P(2ap, ap 2 )
We know the coordinates
of an external point (T)
Q(2aq, aq 2 ) From this external point,
x two tangents can be drawn
meeting the parabola at P
T  x0 , y0  and Q.

7. Chord of Contact
y x 2  4ay
P(2ap, ap 2 )
We know the coordinates
of an external point (T)
Q(2aq, aq 2 ) From this external point,
x two tangents can be drawn
meeting the parabola at P
T  x0 , y0  and Q.

8. Chord of Contact
y x 2  4ay
P(2ap, ap 2 )
We know the coordinates
of an external point (T)
Q(2aq, aq 2 ) From this external point,
x two tangents can be drawn
meeting the parabola at P
T  x0 , y0  and Q.
The line joining these two
points is called the chord
of contact.

9. Chord of Contact
y x 2  4ay
P(2ap, ap 2 )
We know the coordinates
of an external point (T)
Q(2aq, aq 2 ) From this external point,
x two tangents can be drawn
meeting the parabola at P
T  x0 , y0  and Q.
The line joining these two
points is called the chord
of contact.

10. (1) Parametric approach

11. (1) Parametric approach
1 Show that PQ has equation  p  q  x  2 y  2apq

12. (1) Parametric approach
1 Show that PQ has equation  p  q  x  2 y  2apq
2 Show the two tangents have equations
px  y  ap 2  0 and qx  y  aq 2  0

13. (1) Parametric approach
1 Show that PQ has equation  p  q  x  2 y  2apq
2 Show the two tangents have equations
px  y  ap 2  0 and qx  y  aq 2  0
3 Show that T is the point a  p  q  , apq

14. (1) Parametric approach
1 Show that PQ has equation  p  q  x  2 y  2apq
2 Show the two tangents have equations
px  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 

15. (1) Parametric approach
1 Show that PQ has equation  p  q  x  2 y  2apq
2 Show the two tangents have equations
px  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 
x0
pq 
a

16. (1) Parametric approach
1 Show that PQ has equation  p  q  x  2 y  2apq
2 Show the two tangents have equations
px  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  apq
x0
pq 
a

17. (1) Parametric approach
1 Show that PQ has equation  p  q  x  2 y  2apq
2 Show the two tangents have equations
px  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  apq
x0
pq 
a
x0 x
PQ is  2 y  2 y0
a

18. (1) Parametric approach
1 Show that PQ has equation  p  q  x  2 y  2apq
2 Show the two tangents have equations
px  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  apq
x0
pq 
a
x0 x
PQ is  2 y  2 y0
a
Hence the chord of contact is x0 x  2a y0  y 

19. (1) Parametric approach
1 Show that PQ has equation  p  q  x  2 y  2apq
2 Show the two tangents have equations
px  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  apq
x0
pq 
a
x0 x notice
PQ is  2 y  2 y0
a similarity
to tangent
Hence the chord of contact is x0 x  2a y0  y 

20. (2) Cartesian approach

21. (2) Cartesian approach
1 Show that PT has equation xx1  2a  y  y1 

22. (2) Cartesian approach
1 Show that PT has equation xx1  2a  y  y1 
T lies on PT  x0 x1  2a y0  y1 

23. (2) Cartesian approach
1 Show 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 

24. (2) Cartesian approach
1 Show 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 

25. (2) Cartesian approach
1 Show 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 

26. (2) Cartesian approach
1 Show 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 

27. (2) Cartesian approach
1 Show 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 

28. (2) Cartesian approach
1 Show 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