The document discusses chords of a parabola. It defines a chord as a line segment connecting two points on a parabola. It shows that the slope of any chord can be expressed as a simple formula involving the x-intercepts of the chord points. It also proves that the slope of a focal chord, which passes through the focus, is always 1/2.

1. Chords of a Parabola

2. Chords of a Parabola
(1) Chord

3. Chords of a Parabola
(1) Chord
y x 2  4ay
x

4. Chords of a Parabola
(1) Chord
y x 2  4ay
P(2ap, ap 2 )
x

5. Chords of a Parabola
(1) Chord
y x 2  4ay
P(2ap, ap 2 )
Q(2aq, aq 2 )
x

6. Chords of a Parabola
(1) Chord
y x 2  4ay
P(2ap, ap 2 )
Q(2aq, aq 2 )
x

7. Chords of a Parabola
(1) Chord
y x 2  4ay
P(2ap, ap 2 )
Q(2aq, aq 2 )
x
ap 2  aq 2
mPQ 
2ap  2aq

8. Chords of a Parabola
(1) Chord
y x 2  4ay
P(2ap, ap 2 )
Q(2aq, aq 2 )
x
ap 2  aq 2
mPQ 
2ap  2aq
a p  q  p  q 

2a  p  q 
pq

2

9. Chords of a Parabola
(1) Chord
y x 2  4ay
P(2ap, ap 2 )
Q(2aq, aq 2 )
x
ap 2  aq 2 pq
mPQ  y  ap 
2
 x  2ap 
2ap  2aq 2
a p  q  p  q 

2a  p  q 
pq

2

10. Chords of a Parabola
(1) Chord
y x 2  4ay
P(2ap, ap 2 )
Q(2aq, aq 2 )
x
ap 2  aq 2 pq
mPQ  y  ap 
2
 x  2ap 
2ap  2aq 2
a p  q  p  q 
 2 y  2ap 2   p  q x  2ap 2  2apq
2a  p  q 
pq

2

11. Chords of a Parabola
(1) Chord
y x 2  4ay
P(2ap, ap 2 )
Q(2aq, aq 2 )
x
ap 2  aq 2 pq
mPQ  y  ap 
2
 x  2ap 
2ap  2aq 2
a p  q  p  q 
 2 y  2ap 2   p  q x  2ap 2  2apq
2a  p  q 
pq  p  q x  2 y  2apq

2

12. (2) Focal Chord (chord passes through focus)

13. (2) Focal Chord (chord passes through focus)
y x 2  4ay
P(2ap, ap 2 ) Q(2aq, aq 2 )
x

14. (2) Focal Chord (chord passes through focus)
y x 2  4ay
S  0, a 
2
P(2ap, ap ) Q(2aq, aq 2 )
x

15. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
S  0, a 
2
P(2ap, ap ) Q(2aq, aq 2 )
x

16. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
ap  a
2
2 mPS 
S  0, a  2ap  0
P(2ap, ap 2 ) Q(2aq, aq 2 )
x

17. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
ap  a
2
2 mPS 
S  0, a  2ap  0
P(2ap, ap 2 )
a  p 2  1
Q(2aq, aq 2 )

x 2ap
p2 1

2p

18. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
ap  a
2
2 mPS 
S  0, a  2ap  0
P(2ap, ap 2 )
a  p 2  1
Q(2aq, aq 2 )

x 2ap
p2 1

2p
If PQ is a focal chord
mPQ  mPS

19. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
ap  a
2
2 mPS 
S  0, a  2ap  0
P(2ap, ap 2 )
a  p 2  1
Q(2aq, aq 2 )

x 2ap
p2 1

2p
If PQ is a focal chord
mPQ  mPS
p2 1 p  q

2p 2

20. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
ap  a
2
2 mPS 
S  0, a  2ap  0
P(2ap, ap 2 )
a  p 2  1
Q(2aq, aq 2 )

x 2ap
p2 1

2p
If PQ is a focal chord
mPQ  mPS
p2 1 p  q

2p 2
p 2  1  p 2  pq
pq  1

21. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
ap  a
2
2 mPS 
S  0, a  2ap  0
P(2ap, ap 2 )
a  p 2  1
Q(2aq, aq 2 )

x 2ap
p2 1

2p
OR If PQ is a focal chord
(if you have derived equation of chord already) mPQ  mPS
p2 1 p  q

2p 2
p 2  1  p 2  pq
pq  1

22. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
ap  a
2
2 mPS 
S  0, a  2ap  0
P(2ap, ap 2 )
a  p 2  1
Q(2aq, aq 2 )

x 2ap
p2 1

2p
OR If PQ is a focal chord
(if you have derived equation of chord already) mPQ  mPS
1  p  q x  2 y  2apq p2 1 p  q

2p 2
p 2  1  p 2  pq
pq  1

23. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
ap  a
2
2 mPS 
S  0, a  2ap  0
P(2ap, ap 2 )
a  p 2  1
Q(2aq, aq 2 )

x 2ap
p2 1

2p
OR If PQ is a focal chord
(if you have derived equation of chord already) mPQ  mPS
1  p  q x  2 y  2apq p2 1 p  q

2 (0,a) lies on the chord: 2p 2
p 2  1  p 2  pq
pq  1

24. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
ap  a
2
2 mPS 
S  0, a  2ap  0
P(2ap, ap 2 )
a  p 2  1
Q(2aq, aq 2 )

x 2ap
p2 1

2p
OR If PQ is a focal chord
(if you have derived equation of chord already) mPQ  mPS
1  p  q x  2 y  2apq p2 1 p  q

2 (0,a) lies on the chord:  2a  2apq 2p 2
pq  1 p 2  1  p 2  pq
pq  1

25. (2) Focal Chord (chord passes through focus)
y x 2  4ay pq
1 Prove mPQ 
2
ap  a
2
2 mPS 
S  0, a  2ap  0
P(2ap, ap 2 )
a  p 2  1
Q(2aq, aq 2 )

x 2ap
p2 1

2p
OR If PQ is a focal chord
(if you have derived equation of chord already) mPQ  mPS
1  p  q x  2 y  2apq p2 1 p  q

2 (0,a) lies on the chord:  2a  2apq 2p 2
pq  1 p 2  1  p 2  pq
Exercise 9E; 1ac, 2, 3, 4, 5, 7 pq  1