x2 t03 06 chord of contact & properties (2013)
TRANSCRIPT
Chord of Contacty
xa-a
b
-b
00 , yxT
From an external point T, two tangents may be drawn.
Chord of Contact
1equation has at tangent 21
21
byy
axxP
y
xa-a
b
-b
11, yxP 00 , yxT
From an external point T, two tangents may be drawn.
Chord of Contact
1equation has at tangent 21
21
byy
axxP
y
xa-a
b
-b
11, yxP
1equation has at tangent 22
22
byy
axxQ
22 , yxQ
00 , yxT
From an external point T, two tangents may be drawn.
Chord of Contact
1equation has at tangent 21
21
byy
axxP
y
xa-a
b
-b
11, yxP
1equation has at tangent 22
22
byy
axxQ
22 , yxQ
Now T
00 , yxT
lies on both lines,
From an external point T, two tangents may be drawn.
Chord of Contact
1equation has at tangent 21
21
byy
axxP
y
xa-a
b
-b
11, yxP
1equation has at tangent 22
22
byy
axxQ
22 , yxQ
Now T
00 , yxT
lies on both lines,1 and 1 2
022
022
012
01 b
yyaxx
byy
axx
From an external point T, two tangents may be drawn.
y
xa-a
b
-b
11, yxP
22 , yxQ
00 , yxT
1201
201
byy
axx
1202
202
byy
axx
Thus P and Q both must lie on a line with equation;
y
xa-a
b
-b
11, yxP
22 , yxQ
00 , yxT
1201
201
byy
axx
1202
202
byy
axx
Thus P and Q both must lie on a line with equation;
120
20
byy
axx
y
xa-a
b
-b
11, yxP
22 , yxQ
00 , yxT
1201
201
byy
axx
1202
202
byy
axx
Thus P and Q both must lie on a line with equation;
120
20
byy
axx
which must be the line PQ i.e. chord of contact
y
xa-a
b
-b
11, yxP
22 , yxQ
00 , yxT
1201
201
byy
axx
1202
202
byy
axx
Thus P and Q both must lie on a line with equation;
120
20
byy
axx
which must be the line PQ i.e. chord of contact
Similarly the chord of contact of the hyperbola has the equation;
120
20
byy
axx
Rectangular Hyperbola y
x2cxy
1) Show that equation PQ
pccpP ,
qccqQ ,
)1(is qpcpqyx
Rectangular Hyperbola y
x2cxy
1) Show that equation PQ
pccpP ,
qccqQ ,
)1(is qpcpqyx
2) Show that T
00 , yxT
qpc
qpcpq 2,2 scoordinate has
Rectangular Hyperbola y
x2cxy
1) Show that equation PQ
pccpP ,
qccqQ ,
)1(is qpcpqyx
2) Show that T
00 , yxT
qpc
qpcpq 2,2 scoordinate has
qpcy
qpcpqx
2 200
Rectangular Hyperbola y
x2cxy
1) Show that equation PQ
pccpP ,
qccqQ ,
)1(is qpcpqyx
2) Show that T
00 , yxT
qpc
qpcpq 2,2 scoordinate has
qpcy
qpcpqx
2 200
Substituting into (1); qpcyc
xqpx
2
0
Rectangular Hyperbola y
x2cxy
1) Show that equation PQ
pccpP ,
qccqQ ,
)1(is qpcpqyx
2) Show that T
00 , yxT
qpc
qpcpq 2,2 scoordinate has
qpcy
qpcpqx
2 200
Substituting into (1); qpcyc
xqpx
2
0
0
2
0
0 222
ycy
cycxx
Rectangular Hyperbola y
x2cxy
1) Show that equation PQ
pccpP ,
qccqQ ,
)1(is qpcpqyx
2) Show that T
00 , yxT
qpc
qpcpq 2,2 scoordinate has
qpcy
qpcpqx
2 200
Substituting into (1); qpcyc
xqpx
2
0
0
2
0
0 222
ycy
cycxx 2
00 2cyxxy
Geometric Properties
Geometric Properties(1) The chord of contact from a point on the directrix is a focal
chord.
Geometric Properties(1) The chord of contact from a point on the directrix is a focal
chord.
ellipse
0, scoordinate hasit directrix on the is As yeaT
eax 0 i.e.
Geometric Properties(1) The chord of contact from a point on the directrix is a focal
chord.
ellipse
0, scoordinate hasit directrix on the is As yeaT
eax 0 i.e.
equation;thehavellcontact wi of chord
1
1
20
20
2
byy
aex
byy
aeax
Geometric Properties(1) The chord of contact from a point on the directrix is a focal
chord.
ellipse
0, scoordinate hasit directrix on the is As yeaT
eax 0 i.e.
equation;thehavellcontact wi of chord
1
1
20
20
2
byy
aex
byy
aeax
Substitute in focus (ae,0)
Geometric Properties(1) The chord of contact from a point on the directrix is a focal
chord.
ellipse
0, scoordinate hasit directrix on the is As yeaT
eax 0 i.e.
equation;thehavellcontact wi of chord
1
1
20
20
2
byy
aex
byy
aeax
Substitute in focus (ae,0)
1
010
aeae
Geometric Properties(1) The chord of contact from a point on the directrix is a focal
chord.
ellipse
0, scoordinate hasit directrix on the is As yeaT
eax 0 i.e.
equation;thehavellcontact wi of chord
1
1
20
20
2
byy
aex
byy
aeax
Substitute in focus (ae,0)
1
010
aeae
contactofchordon liesfocusi.e. it is a focal chord
(2) That part of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus.
y
x
sin,cos baPT
eax
S
(2) That part of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus.
y
x
sin,cos baPT
eax
SProve: 90PST
(2) That part of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus.
y
x
sin,cos baPT
eax
SProve: 90PST
1sincos is tangent ofequation b
ya
x
1sincos,when
by
aea
eax
(2) That part of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus.
y
x
sin,cos baPT
eax
SProve: 90PST
1sincos is tangent ofequation b
ya
x
1sincos,when
by
aea
eax
sincos
cossin
1sincos
eeby
ee
by
by
e
(2) That part of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus.
y
x
sin,cos baPT
eax
SProve: 90PST
1sincos is tangent ofequation b
ya
x
1sincos,when
by
aea
eax
sincos
cossin
1sincos
eeby
ee
by
by
e
sincos,
eeb
eaT
aeabmPS
cos0sin
eab
cossin
aeabmPS
cos0sin
eab
cossin
aeea
eeb
mTS
0
sincos
2sin
cosaeae
eeb
aeabmPS
cos0sin
eab
cossin
aeea
eeb
mTS
0
sincos
2sin
cosaeae
eeb
sin1
cos2ea
eb
aeabmPS
cos0sin
eab
cossin
aeea
eeb
mTS
0
sincos
2sin
cosaeae
eeb
sin1
cos2ea
eb
sin1cos22
aea
eb
aeabmPS
cos0sin
eab
cossin
aeea
eeb
mTS
0
sincos
2sin
cosaeae
eeb
sin1
cos2ea
eb
sin1cos22
aea
eb
sincos
bea
aeabmPS
cos0sin
eab
cossin
aeea
eeb
mTS
0
sincos
2sin
cosaeae
eeb
sin1
cos2ea
eb
sin1cos22
aea
eb
sincos
bea
1sin
coscos
sin
bea
eabmm TSPS
aeabmPS
cos0sin
eab
cossin
aeea
eeb
mTS
0
sincos
2sin
cosaeae
eeb
sin1
cos2ea
eb
sin1cos22
aea
eb
sincos
bea
1sin
coscos
sin
bea
eabmm TSPS 90 PST
(3) Reflection PropertyTangent to an ellipse at a point P on it is equally inclined to the focal chords through P.
y
x
PT
S
T
S
(3) Reflection PropertyTangent to an ellipse at a point P on it is equally inclined to the focal chords through P.
y
x
PT
S
T
S
Prove: TPSSPT
(3) Reflection PropertyTangent to an ellipse at a point P on it is equally inclined to the focal chords through P.
y
x
PT
S
T
S
Prove: TPSSPT Construct a line || y axis passing through P
(3) Reflection PropertyTangent to an ellipse at a point P on it is equally inclined to the focal chords through P.
y
x
PT
S
T
S
Prove: TPSSPT Construct a line || y axis passing through P
NPPN
TPPT
(ratio of intercepts of || lines)
NN
(3) Reflection PropertyTangent to an ellipse at a point P on it is equally inclined to the focal chords through P.
y
x
PT
S
T
S
Prove: TPSSPT Construct a line || y axis passing through P
NPPN
TPPT
(ratio of intercepts of || lines)
NPTP
PNPT
NN
SPNePPSePN and
SPNePPSePN and
eSPTP
ePSPT
SPNePPSePN and
eSPTP
ePSPT
SPTP
PSPT
SPNePPSePN and
eSPTP
ePSPT
SPTP
PSPT
90 TSPPST (proven in property (2))
SPNePPSePN and
eSPTP
ePSPT
SPTP
PSPT
90 TSPPST (proven in property (2))
TPSSPT secsec
SPNePPSePN and
eSPTP
ePSPT
SPTP
PSPT
90 TSPPST (proven in property (2))
TPSSPT secsec
TPSSPT
SPNePPSePN and
eSPTP
ePSPT
SPTP
PSPT
90 TSPPST (proven in property (2))
TPSSPT secsec
TPSSPT
e.g. Find the cartesian equation of 2 2 8z z
e.g. Find the cartesian equation of 2 2 8z z
The sum of the focal lengths of an ellipse is constant
e.g. Find the cartesian equation of 2 2 8z z
2 84
aa
The sum of the focal lengths of an ellipse is constant
e.g. Find the cartesian equation of 2 2 8z z
2 84
aa
2ae
4 212
e
e
The sum of the focal lengths of an ellipse is constant
e.g. Find the cartesian equation of 2 2 8z z
2 84
aa
2ae
4 212
e
e
2 2 21b a e
2 116 14
12
b
The sum of the focal lengths of an ellipse is constant
e.g. Find the cartesian equation of 2 2 8z z
2 84
aa
2ae
4 212
e
e
2 2 21b a e
2 116 14
12
b
The sum of the focal lengths of an ellipse is constant
2 2
locus is the ellipse 116 12x y
e.g. Find the cartesian equation of 2 2 8z z
2 84
aa
2ae
4 212
e
e
2 2 21b a e
2 116 14
12
b
Exercise 6E; 1, 2, 4, 7, 8, 10Exercise 4N; 1 l,m,nKomarami Unit 4
The sum of the focal lengths of an ellipse is constant
2 2
locus is the ellipse 116 12x y