1. Chord of Contact y
b
T  x0 , y0 
-a a x
-b
From an external point T, two tangents may be drawn.

2. Chord of Contact
y
b
P x1 , y1 
T  x0 , y0 
-a a x
-b
From an external point T, two tangents may be drawn.
xx yy
tangent at P has equation 12  12  1
a b

3. Chord of Contact
y
b
P x1 , y1 
T  x0 , y0 
-a a x
Q  x2 , y 2 
-b
From an external point T, two tangents may be drawn.
xx yy
tangent at P has equation 12  12  1
a b
xx y y
tangent at Q has equation 22  22  1
a b

4. Chord of Contact
y
b
P x1 , y1 
T  x0 , y0 
-a a x
Q  x2 , y 2 
-b
From an external point T, two tangents may be drawn.
xx yy
tangent at P has equation 12  12  1
a b
xx y y
tangent at Q has equation 22  22  1
Now T lies on both lines, a b

5. Chord of Contact
y
b
P x1 , y1 
T  x0 , y0 
-a a x
Q  x2 , y 2 
-b
From an external point T, two tangents may be drawn.
xx yy
tangent at P has equation 12  12  1
a b
xx y y
tangent at Q has equation 22  22  1
Now T lies on both lines, a b
xx yy x2 x0 y2 y0
 1 2 0  1 2 0  1 and 2
 2 1
a b a b

6. y
x2 x0 y2 y0
b  2 1
P x1 , y1  a
2
b
T  x0 , y0 
-a a x
Q  x2 , y 2  x1 x0 y1 y0
 2 1
2
-b a b
Thus P and Q both must lie on a line with equation;

7. y
x2 x0 y2 y0
b  2 1
P x1 , y1  a
2
b
T  x0 , y0 
-a a x
Q  x2 , y 2  x1 x0 y1 y0
 2 1
2
-b a b
Thus P and Q both must lie on a line with equation;
x0 x y0 y
2
 2 1
a b

8. y
x2 x0 y2 y0
b  2 1
P x1 , y1  a
2
b
T  x0 , y0 
-a a x
Q  x2 , y 2  x1 x0 y1 y0
 2 1
2
-b a b
Thus P and Q both must lie on a line with equation;
x0 x y0 y
2
 2 1
a b
which must be the line PQ i.e. chord of contact

9. y
x2 x0 y2 y0
b  2 1
P x1 , y1  a
2
b
T  x0 , y0 
-a a x
Q  x2 , y 2  x1 x0 y1 y0
 2 1
2
-b a b
Thus P and Q both must lie on a line with equation;
x0 x y0 y
2
 2 1
a b
which must be the line PQ i.e. chord of contact
Similarly the chord of contact of the hyperbola has the equation;
x0 x y0 y
2
 2 1
a b

10. Rectangular Hyperbola y
 c
P cp, 
 p  c
Q cq, 
 q
x
xy  c 2
1) Show that equation PQ is x  pqy  c p  q  (1)

11. Rectangular Hyperbola y
 c
P cp, 
 p  c
Q cq, 
T  x0 , y0   q
x
xy  c 2
1) Show that equation PQ is x  pqy  c p  q  (1)
 2cpq 2c 
2) Show that T has coordinates  , 
 p  q p  q

12. Rectangular Hyperbola y
 c
P cp, 
 p  c
Q cq, 
T  x0 , y0   q
x
xy  c 2
1) Show that equation PQ is x  pqy  c p  q  (1)
 2cpq 2c 
2) Show that T has coordinates  , 
 p  q p  q
2cpq 2c
 x0  y0 
pq pq

13. Rectangular Hyperbola y
 c
P cp, 
 p  c
Q cq, 
T  x0 , y0   q
x
xy  c 2
1) Show that equation PQ is x  pqy  c p  q  (1)
 2cpq 2c 
2) Show that T has coordinates  , 
 p  q p  q
2cpq 2c
 x0  y0 
pq pq
Substituting into (1); x   p  q x0 y  c p  q 
2c

14. Rectangular Hyperbola y
 c
P cp, 
 p  c
Q cq, 
T  x0 , y0   q
x
xy  c 2
1) Show that equation PQ is x  pqy  c p  q  (1)
 2cpq 2c 
2) Show that T has coordinates  , 
 p  q p  q
2cpq 2c
 x0  y0 
pq pq
Substituting into (1); x   p  q x0 y  c p  q 
2c
2cx0 2c 2
x y
2cy0 y0

15. Rectangular Hyperbola y
 c
P cp, 
 p  c
Q cq, 
T  x0 , y0   q
x
xy  c 2
1) Show that equation PQ is x  pqy  c p  q  (1)
 2cpq 2c 
2) Show that T has coordinates  , 
 p  q p  q
2cpq 2c
 x0  y0 
pq pq
Substituting into (1); x   p  q x0 y  c p  q 
2c
2cx0 2c 2
x y xy0  x0 y  2c 2
2cy0 y0

16. Geometric Properties

17. Geometric Properties
(1) The chord of contact from a point on the directrix is a focal
chord.

18. Geometric Properties
(1) The chord of contact from a point on the directrix is a focal
chord.
ellipse
 a , y  i.e. x  a
As T is on the directrix it has coordinates  0 0
 e  e

19. Geometric Properties
(1) The chord of contact from a point on the directrix is a focal
chord.
ellipse
 a , y  i.e. x  a
As T is on the directrix it has coordinates  0 0
 e  e
 chord of contact will have the equation; a
x 
 e   yy0  1
a2 b2
x yy0
 2 1
ae b

20. Geometric Properties
(1) The chord of contact from a point on the directrix is a focal
chord.
ellipse
 a , y  i.e. x  a
As T is on the directrix it has coordinates  0 0
 e  e
 chord of contact will have the equation; a
x 
 e   yy0  1
a2 b2
x yy0
 2 1
Substitute in focus (ae,0) ae b

21. Geometric Properties
(1) The chord of contact from a point on the directrix is a focal
chord.
ellipse
 a , y  i.e. x  a
As T is on the directrix it has coordinates  0 0
 e  e
 chord of contact will have the equation; a
x 
 e   yy0  1
a2 b2
x yy0
 2 1
Substitute in focus (ae,0) ae b
ae
 0  1 0
ae
1

22. Geometric Properties
(1) The chord of contact from a point on the directrix is a focal
chord.
ellipse
As T is on the directrix it has coordinates   a , y  i.e. x  a
0 0
 e  e
 chord of contact will have the equation; a
x 
 e   yy0  1
a2 b2
x yy0
 2 1
Substitute in focus (ae,0) ae b
ae
 0  1 0  focus lies on chord of contact
ae
i.e. it is a focal chord
1

23. (2) That part of the tangent between the point of contact and the
directrix subtends a right angle at the corresponding focus.
y
Pa cos , b sin  
T
S x
a
x
e

24. (2) That part of the tangent between the point of contact and the
directrix subtends a right angle at the corresponding focus.
y
Pa cos , b sin  
T
Prove: PST  90 S x
a
x
e

25. (2) That part of the tangent between the point of contact and the
directrix subtends a right angle at the corresponding focus.
y
Pa cos , b sin  
T
Prove: PST  90 S x
x cos y sin 
equation of tangent is  1 a
a a b x
cos e
a y sin 
when x  , e  1
e a b

26. (2) That part of the tangent between the point of contact and the
directrix subtends a right angle at the corresponding focus.
y
Pa cos , b sin  
T
Prove: PST  90 S x
x cos y sin 
equation of tangent is  1 a
a a b x
cos e
a y sin 
when x  , e  1
e a b
cos y sin 
 1
e b
y sin  e  cos

b e
be  cos 
y
e sin 

27. (2) That part of the tangent between the point of contact and the
directrix subtends a right angle at the corresponding focus.
y
Pa cos , b sin  
T
Prove: PST  90 S x
x cos y sin 
equation of tangent is  1 a
a a b x
cos e
a y sin 
when x  , e  1
e a b
cos y sin 
 1
e b
y sin  e  cos a be  cos  
 T  ,
 
b e e e sin  
be  cos 
y
e sin 

28. b sin   0
mPS 
a cos  ae
b sin 

acos  e 

29. b sin   0 be  cos 
mPS  0
a cos  ae mTS  e sin 
b sin  a
 ae

acos  e  e
be  cos  e
 
e sin  a  ae 2

30. b sin   0 be  cos 
mPS  0
a cos  ae mTS  e sin 
b sin  a
 ae

acos  e  e
be  cos  e
 
e sin  a  ae 2
be  cos 

a 1  e 2 sin 

31. b sin   0 be  cos 
mPS  0
a cos  ae mTS  e sin 
b sin  a
 ae

acos  e  e
be  cos  e
 
e sin  a  ae 2
be  cos 

a 1  e 2 sin 
be  cos 
 2
a 1  e 2 
sin 
a

32. b sin   0 be  cos 
mPS  0
a cos  ae mTS  e sin 
b sin  a
 ae

acos  e  e
be  cos  e
 
e sin  a  ae 2
be  cos 

a 1  e 2 sin 
be  cos 
 2
a 1  e 2 
sin 
a
ae  cos 

b sin 

33. b sin   0 be  cos 
mPS  0
a cos  ae mTS  e sin 
b sin  a
 ae

acos  e  e
be  cos  e
 
e sin  a  ae 2
be  cos 

a 1  e 2 sin 
be  cos 
 2
a 1  e 2 
sin 
a
ae  cos 

b sin 
b sin  ae  cos 
mPS  mTS  
acos  e  b sin 
 1

34. b sin   0 be  cos 
mPS  0
a cos  ae mTS  e sin 
b sin  a
 ae

acos  e  e
be  cos  e
 
e sin  a  ae 2
be  cos 

a 1  e 2 sin 
be  cos 
 2
a 1  e 2 
sin 
a
ae  cos 

b sin 
b sin  ae  cos 
mPS  mTS  
acos  e  b sin   PST  90
 1

35. (3) Reflection Property
Tangent to an ellipse at a point P on it is equally inclined to
the focal chords through P.
T y
P
T
S S x

36. (3) Reflection Property
Tangent to an ellipse at a point P on it is equally inclined to
the focal chords through P.
T y
P
T
S S x
Prove: SPT  S PT 

37. (3) Reflection Property
Tangent to an ellipse at a point P on it is equally inclined to
the focal chords through P.
T y
P
T
S S x
Prove: SPT  S PT 
Construct a line || y axis passing through P

38. (3) Reflection Property
Tangent to an ellipse at a point P on it is equally inclined to
the focal chords through P.
T y
P
N N
T
S S x
Prove: SPT  S PT 
Construct a line || y axis passing through P
PT PN
 (ratio of intercepts of || lines)
PT  PN 

39. (3) Reflection Property
Tangent to an ellipse at a point P on it is equally inclined to
the focal chords through P.
T y
P
N N
T
S S x
Prove: SPT  S PT 
Construct a line || y axis passing through P
PT PN
 (ratio of intercepts of || lines)
PT  PN 
PT PT 
 
PN PN 

40. ePN  PS and ePN   PS 

41. ePN  PS and ePN   PS 
PT PT 
 
PS PS 
e e

42. ePN  PS and ePN   PS 
PT PT 
 
PS PS 
e e
PT PT 

PS PS 

43. ePN  PS and ePN   PS 
PT PT 
 
PS PS 
e e
PT PT 

PS PS 
PST  PS T   90 (proven in property (2))

44. ePN  PS and ePN   PS 
PT PT 
 
PS PS 
e e
PT PT 

PS PS 
PST  PS T   90 (proven in property (2))
 sec SPT  sec S PT 

45. ePN  PS and ePN   PS 
PT PT 
 
PS PS 
e e
PT PT 

PS PS 
PST  PS T   90 (proven in property (2))
 sec SPT  sec S PT 
SPT  S PT 

46. ePN  PS and ePN   PS 
PT PT 
 
PS PS 
e e
PT PT 

PS PS 
PST  PS T   90 (proven in property (2))
 sec SPT  sec S PT 
SPT  S PT 

47. e.g. Find the cartesian equation of z  2  z  2  8

48. e.g. Find the cartesian equation of z  2  z  2  8
The sum of the focal lengths of an ellipse is constant

49. e.g. Find the cartesian equation of z  2  z  2  8
The sum of the focal lengths of an ellipse is constant
2a  8
a4

50. e.g. Find the cartesian equation of z  2  z  2  8
The sum of the focal lengths of an ellipse is constant
2a  8 ae  2
a4 4e  2
1
e
2

51. e.g. Find the cartesian equation of z  2  z  2  8
The sum of the focal lengths of an ellipse is constant
2a  8 ae  2 b 2  a 2 1  e 2 
a4 4e  2  1
b 2  16 1  
e
1  4
2  12

52. e.g. Find the cartesian equation of z  2  z  2  8
The sum of the focal lengths of an ellipse is constant
2a  8 ae  2 b 2  a 2 1  e 2 
a4 4e  2  1
b 2  16 1  
e
1  4
2  12
x2 y2
 locus is the ellipse  1
16 12

53. e.g. Find the cartesian equation of z  2  z  2  8
The sum of the focal lengths of an ellipse is constant
2a  8 ae  2 b 2  a 2 1  e 2 
a4 4e  2  1
b 2  16 1  
e
1  4
2  12
x2 y2
 locus is the ellipse  1
16 12
Exercise 6E; 1, 2, 4, 7, 8, 10