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FURTHERPURE3SKE
FURTHERCOORDINATESYSTEMS
Philosophy of the course
• The course does not aim to present a teaching plan for any of
the FP3 topics. This is left to your professional judgement as a
teacher.
• The course is more about presenting each topic in a way that
provides insights into the topic beyond the syllabus.
• In general the course will present the necessary knowledge and
skills to teach the Edexcel FP3 module effectively.
• Relevant examination problem solving techniques will be
demonstrated.
Session structure
• Section 1: Introduction to the ellipse and hyperbola.
• Section 2: Cartesian and parametric equations of the ellipse and
hyperbola.
• Section 3: Properties of the ellipse and hyperbola: eccentricity,
foci and directrices.
• Section 4: Tangents and normals to the ellipse and hyperbola.
• Section 5. Typical examination questions.
•Section 1: Introduction to the
ellipse and hyperbola.
We get the hyperbola
and the ellipse.
If a double
cone is cut in
a certain way.
The circle, ellipse, parabola and hyperbola are called the conic
sections. This web page http://bit.ly/1dJzJ1Y gives an animation of
how an ellipse might be generated from a cone.
Essentially this topic extends the study of FP1 Coordinate geometry which
introduced some fundamental properties of the parabola and hyperbola. Here we
study the ellipse and the hyperbola to a greater depth.
The ellipse and the hyperbola are from a family of curves called the conic sections.
• Apollonius of Perga (approx. 262 BC–190 BC) was a Greek geometer who
studied with Euclid. He is best known for his work on cross sections of a cone.
This is the first recorded study of the conic sections and led to a school of
mathematics which developed the important work of Apollonius.
• Important Arabic additions to the conics from the tenth and eleventh century
were also crucial to the mathematics of the scientific revolution.
• In fact, even to this day, large parts The Conics of Apollonius do not exist in
their original Greek form, and are known to us only through Arabic translations.
• http://www.quadrivium.info/MathInt/Notes/Apollonius.pdf
Abrief history of the conic sections.
The importance of the ellipse.
Kepler model of the elliptical orbits of
planets in a solar system →
The ellipse was brought into scientific prominence when Johannes
Kepler proved that planetary orbits were not circular but elliptical.
http://csep10.phys.utk.edu/astr161/lect/history/kepler.html
Keplers laws of planetary motion:
First Law: The orbit of every planet is an ellipse with the sun at one of the
foci.
Second Law: A line joining a planet and the Sun sweeps out equal areas
during equal intervals of time.
Third Law: The square of the orbital period of a planet is directly
proportional to the cube of the semi-major axis of its orbit.
Hyperbolae are also important in astronomy:
The majority of comets are
in hyperbolic or parabolic
orbits.
If a comet’s initial velocity is
insignificant the path is
parabolic, whereas if the
initial velocity is significant,
the path is hyperbolic.
http://bit.ly/1f5Xi21
Ellipses and hyperbolae are also important
in other aspects of science and in
economics. You can find more here:
http://bit.ly/1igiw26
•Section 2: Cartesian and
parametric equations of the
ellipse and hyperbola.
The ellipse as a transformed circle.
a
O
.
,
value
fixed
by the
circle
on the
points
all
of
coordinate
he
Multiply t
:
way
in this
circle
the
transform
Now
a
b
a
b
P
y

:
0
,
circle
he
Consider t 2
2
2


 a
a
y
x
●(x, y)
● 





y
a
b
x,
circle.
on the
lies
)
,
(
then
ellipse
on the
lies
)
,
(
If
:
follows
as
found
be
can
ellipse
the
of
equation
The
Y
b
a
X
Y
X
ellipse.
general
the
of
equation
Cartesian
The
1
So
2
2
2
2
2
2
2











b
Y
a
X
a
Y
b
a
X
red.
in
shown
ellipse,
an
is
circle
ed
transform
The
Fundamental properties of the ellipse.
A: (a, 0)
O
b
y
b
y
x
y
a
x
a
x
y
x
b
y
a
x














1
0
when
axis
the
intersects
1
0
when
axis
the
intersects
1
equation
Cartesian
with
ellipse
The
2
2
2
2
2
2
2
2
B (–a, 0)
C (0, b)
D (0, –b)
If a > b then AB is called the
major axis and CD is called the
minor axis of the ellipse.
If the other case when b < a,
AB will be the minor and CD
the major axis.
Parametric equations of the ellipse.
(a, 0)
O
diagram.
in the
indicated
are
2π
,
2
3π
π,
,
2
π
,
0
with
ellipse
on the
points
The
(1).
equation
Cartesian
with
ellipse
this
of
equations
parametric
the
gives
)
2
(
)
2
(
..........
sin
,
cos
or
sin
,
cos
writing
to
us
leads
This
1
sin
cos
it with
comparing
by
ed
parametris
be
can
)
1
.(
..........
1
equation
Cartesian
with
ellipse
The
2
2
2
2
2
2









t
t
b
y
t
a
x
t
b
y
t
a
x
t
t
b
y
a
x
(0, b)
t = 0
t = 𝟏
𝟐

t = 
t = 3
2

t = 2
Rectangular to general hyperbolae
   








































































hyperbola
ed
transform
the
of
equation
the
:
2
2
are
s
coordinate
new
the
So
matrix
the
by
ed
transform
are
s
coordinate
the
know
we
FP1
From
clockwise.
4
by
graph
this
rotate
we
Now
graph
has
This
FP1.
in
met
hyperbola
r
rectangula
he
Consider t
2
2
2
2
2
2
1
2
2
1
2
2
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
c
Y
X
xy
Y
X
y
x
y
x
Y
X
y
x
y
x
Y
X
y
x
Y
X
c
xy

↓ rotate ¼
clockwise
Note: x = 0 and y = 0
are lines of symmetry
Rectangular to general hyperbolae




 c
Y
X
c
xy 2
2
2
Perpendicular
asymptotes.
Perpendicular
asymptotes.
later.
verified
be
will
This
:
1
is
0
and
0
symmetry
of
lines
with
hyperbolae
general
the
of
equation
Cartesian
The
asymptotes
lar
perpendicu
have
not
do
hyperbolae
general
In
2
2
2
2





b
y
a
x
y
x
Properties and parametric equations of the hyperbola.
)
1
....(
tan
,
sec
s
coordinate
parametric
implies
1
tan
sec
identity
The
).
0
,
(
and
)
0
,
(
at
axis
Intersects
1
2
2
2
2
2
2
t
b
y
t
a
x
t
t
a
a
x
b
y
a
x








(– a, 0) (a, 0)
t = 0
t = 
)
2
....(
sinh
,
cosh
use
we
,
For
.
1
cosh
because
when
applies
only
But this
)
2
....(
sinh
,
cosh
s
coordinate
parametric
implies
1
sinh
cosh
and
1
Also 2
2
2
2
2
2
a
t
b
y
t
a
x
a
x
t
a
x
t
b
y
t
a
x
t
t
b
y
a
x












(a, 0)
t = 0
Exercises: Equations of the ellipse and hyperbola
functions.
tric
trigonome
of
in terms
form
parametric
the
Give
.
graph.
sketch
a
Provide
.
1
form
equivalent
the
Give
:
above
equations
the
of
each
For
144
16
36
.
4
36
9
4
.
3
36
4
9
.
2
16
4
.
1
2
2
2
2
2
2
2
2
2
2
2
2
iii
ii
b
y
a
x
i.
y
x
y
x
y
x
y
x










Exercises: Equations of the ellipse and hyperbola (solutions)
.
sin
2
,
cos
4
:
is
form
parametric
The
.
:
is
graph
sketch
A
.
1
2
4
16
4
16
4
.
1
2
2
2
2
2
2
2
2

 








y
x
iii
ii
y
x
y
x
i.
y
x
4
– 4
– 2
2
.
sin
3
,
cos
2
:
is
form
parametric
The
.
:
is
graph
sketch
A
.
1
3
2
36
4
9
36
4
9
.
2
2
2
2
2
2
2
2
2

 








y
x
iii
ii
y
x
y
x
i.
y
x
2
– 2
– 3
3
.
tan
2
,
sec
3
:
is
form
parametric
The
.
:
is
graph
sketch
A
.
1
2
3
36
9
4
36
9
4
.
3
2
2
2
2
2
2
2
2

 








y
x
iii
ii
y
x
y
x
i.
y
x
3
– 3
.
tan
3
,
sec
2
:
is
form
parametric
The
.
:
is
graph
sketch
A
.
1
3
2
144
16
36
144
16
36
.
4
2
2
2
2
2
2
2
2

 








y
x
iii
ii
y
x
y
x
i.
y
x
2
– 2
•Section 3: Properties of the
ellipse and hyperbola:
eccentricity, foci and directrices.
Geometrical derivation of the ellipse.
•One way (but not the only one) to define the conic sections -
parabola, hyperbola and ellipse – is the following :
•They are the loci of points P in the plane that satisfy the
following condition:
•The distance of P from a fixed point S = e the distance
of P from a fixed line L.
•The fixed point S is called the focus of the conic section.
•The fixed line L is called its directrix.
•The constant multiple e is called its eccentricity.
• e =1 for the parabola, 0 < e < 1 for the ellipse and e > 1 for the
hyperbola.
Focus, directrix and Cartesian equation of the parabola.
● S (a, 0)
L: x = – a
Let the focus be the point S (a, 0)
and the directrix L: x = – a.
Points P (x, y) on the parabola
satisfy: PS = 1PT.
● P (x, y)
T (– a , y)●
Or PS2 = PT2
ax
y
ax
a
x
y
ax
a
x
a
x
y
a
x
4
2
2
)
(
)
(
2
2
2
2
2
2
2
2
2














Focus, directrix and Cartesian equation of the ellipse.
● S (ae, 0)
L: x = a/e
Here we let the focus be the point S
(ae, 0) and the directrix
L: x = a/e, where 0 < e < 1, is the
eccentricity of the ellipse.
Points P (x, y) on the ellipse
satisfy: PS = ePT : 0 < e < 1
● P (x, y) ● T (a/e , y)
Or PS2 = e2PT2
)
1
(
where
,
1
or
1
)
1
(
)
1
(
)
1
(
2
2
)
(
)
(
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
e
a
b
b
y
a
x
e
a
y
a
x
e
a
y
e
x
aex
a
x
e
y
aex
e
a
x
x
e
a
e
y
ae
x

























Focus, directrix and Cartesian equation of the ellipse.
● S (ae, 0)
L: x = a/e
As the equation of the ellipse has even
powers of x and of y it is symmetrical
about both axes.
This symmetry implies that the ellipse
has two foci and two directrices.
The other focus is S′ (–ae, 0) and other
directrix is L′ : x = –a/e.
● P (x, y) ● T (a/e , y)
.
1
0
),
1
(
where
,
1
is
ellipse
the
of
equation
general
The
2
2
2
2
2
2
2





 e
e
a
b
b
y
a
x
L′ : x = –a/e
S′ (–ae, 0) ●
)
0
,
(
and
)
0
,
(
are
intercepts
the
So
.
1
0
)
0
,
(
and
)
0
,
(
are
intercepts
the
So
.
1
0
2
2
2
2
b
b
y
b
y
b
y
x
a
a
x
a
x
a
x
y














(a, 0)
(–a, 0)
(–b, 0)
(b, 0)
Focus, directrix and Cartesian equation of the hyperbola.
● S (ae, 0)
L: x = a/e
Here we let the focus be the point S
(ae, 0) and the directrix
L: x = a/e, where e, e >1, is the
eccentricity of the hyperbola.
Points P (x, y) on the hyperbola
satisfy: PS = ePT : e > 1
● P (x, y)
T (a/e , y)●
Or PS2 = e2PT2
0
ensure
to
:
)
1
(
write
to
need
t we
reason tha
the
Note
)
1
(
where
,
1
or
1
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
2
2
)
(
)
(
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
































b
e
e
a
b
b
y
a
x
e
a
y
a
x
e
a
y
e
x
e
a
y
e
x
aex
a
x
e
y
aex
e
a
x
e
a
x
e
y
ae
x
Focus, directrix and Cartesian equation of the hyperbola.
Again the equation of the hyperbola has
even powers of x and of y it is
symmetrical about both axes.
The symmetry implies that it has two
foci and two directrices.
The other focus is S′ (–ae, 0) and other
directrix is L′ : x = –a/e.
.
1
),
1
(
where
,
1
is
hyperbola
the
of
equation
general
The
2
2
2
2
2
2
2




 e
e
a
b
b
y
a
x
.
intercepts
no
Hence
.
impossible
is
which
,
1
0
)
0
,
(
and
)
0
,
(
are
intercepts
the
So
.
1
0
2
2
2
2
y
b
y
x
a
a
x
a
x
a
x
y











(a, 0)● ● S (ae, 0)
L: x = a/e
T (a/e , y)●
L′ : x = –a/e
S′ (–ae, 0) ●
● P (x, y)
(–a, 0)
●
Example: Focus and directrix of the ellipse.
12
12
)
4
1
1
(
16
),
1
(
As
4
(1)
Then
)
ellipse
the
for
1
0
(
2
1
4
1
2
8
)
2
(
and
)
1
(
)
2
(
..........
2
So
).
0
,
(
0)
(2,
focus
The
)
1
(
..........
8
So
.
8
directrix
The
:
2
2
2
2
2






















b
b
e
a
b
a
e
e
e
e
e
e
a
ae
e
a
e
a
x
Solution
Example: Focus and directrix of the hyperbola.
5
16
are
s
directrice
The
)
0
,
5
(
)
0
,
(
are
foci
The
hyperbola)
for the
1
(
4
5
16
25
16
9
)
1
(
)
1
(
16
9
So
)
1
(
hyperbola
For the
.
3
and
4
get
we
1
form
standard
with the
1
9
16
Comparing
.
directices
and
foci
the
showing
hyperbola
Sketch the
s.
directrice
its
of
equations
the
and
foci
its
of
s
coordinate
the
Find
1
9
16
equation
has
hyperbola
A
2
2
2
2
2
2
2
2
2
2
2
2
2
2



























e
a
x
ae
e
e
e
e
e
e
a
b
b
a
b
y
a
x
y
x
Solution:
y
x
●
S (5, 0)
S ′(– 5, 0)
●
x = 3.2
x = – 3.2
Exercises: Focus and directrix
above.
sections
conic
the
of
each
for
s
directrice
and
foci
the
Find
1
7
16
.
3
1
3
4
.
2
1
3
4
.
1
2
2
2
2
2
2






y
x
y
x
y
x
Exercises: Focus and directrix (solutions)
3
16
s
Directrice
);
0
,
3
(
)
0
,
(
Foci
.
4
3
.
16
9
)
1
(
16
7
)
1
(
.
7
,
16
ellipse.
An
.
1
7
16
.
3
7
7
4
7
4
s
Directrice
);
0
,
7
(
)
0
,
(
Foci
.
2
7
.
4
7
)
1
(
4
3
so
),
1
(
Here
.
3
,
4
1
:
hyperbola
a
of
equation
the
is
This
.
1
3
4
.
2
4
s
Directrice
);
0
,
1
(
)
0
,
(
Foci
.
2
1
.
4
1
)
1
(
4
3
so
),
1
(
Here
.
3
,
4
1
:
ellipse
an
of
equation
the
is
This
.
1
3
4
.
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2


































































e
a
x
ae
e
e
e
e
a
b
b
a
y
x
e
a
x
ae
e
e
e
e
a
b
b
a
b
y
a
x
y
x
e
a
x
ae
e
e
e
e
a
b
b
a
b
y
a
x
y
x
•Section 4: Tangents and
normals to the ellipse and
hyperbola.
The gradient function of the ellipse.
(1)}
in
s
coordinate
parametric
the
subs
by
(1)
from
derived
be
can
(2)
{n.b.
)
2
.........(
sin
3
cos
2
cos
2
,
sin
2
sin
2
,
cos
3
are
1
4
9
of
s
coordinate
parametric
The
:
ation
differenti
parametric
By
2.
)
1
(
..........
9
4
0
9
4
0
4
2
9
2
1
4
9
:
ation
differenti
implicit
By
1.
:
ways
in two
determined
be
can
function
gradient
The
.
1
4
9
ellipse
he
Consider t
2
2
2
2
2
2
t
t
dx
dy
dt
dx
dt
dy
dx
dy
t
dt
dy
t
dt
dx
t
y
t
x
y
x
y
x
dx
dy
dx
dy
y
x
dx
dy
y
x
y
x
y
x


























The gradient function of the ellipse generalised.
)
2
.........(
sin
cos
)
1
(
..........
.
1
ellipse
For the
:
general
made
be
easily
can
slide
previous
the
of
n
calculatio
The
2
2
2
2
2
2
t
a
t
b
dx
dy
y
a
x
b
dx
dy
b
y
a
x






The equations of the tangent and normal of the ellipse.
before)
as
(
3
2
sin
3
cos
2
sin
3
cos
2
tangent
the
of
gradient
Then the
n
calculatio
or
inspection
By
).
,
(
)
sin
2
,
cos
3
(
in
find
to
need
tion we
differenta
parametric
using
gradient
the
find
To
.
n.b
0
cos
6
sin
6
2
3
or
cos
6
2
sin
6
3
)
cos
3
(
2
)
sin
2
(
3
)
cos
3
(
3
2
)
sin
2
(
:
is
tangent
the
of
equation
an
So
)
sin
2
,
cos
3
(
are
s
coordinate
parametric
The
:
form
parametric
In
2.
0
2
6
2
3
2
3
)
(
2
)
(
3
)
(
3
2
)
(
:
is
tangent
the
of
equation
an
So
3
2
9
4
9
4
tangent
the
of
gradient
The
:
form
Cartesian
In
1.
:
ways
in two
determined
be
can
)
,
(
at
ellipse
this
o
tangent t
the
of
equation
An
.
1
4
9
ellipse
on the
)
,
(
point
he
Consider t
4
4
4
2
2
2
2
2
3
2
2
6
2
2
6
2
2
3
2
2
2
2
2
3
2
2
2
2
2
2
2
2
3
2
2
2
2
2
3
2
2
2
2
2
2
2
3



























































t
t
dx
dy
t
t
t
t
t
t
x
y
t
x
t
y
t
x
t
y
t
x
t
y
t
t
x
y
x
y
x
y
x
y
y
x
dx
dy
y
x
The gradient function of the hyperbola
(1)}
in
s
coordinate
parametric
relevant
the
subs
by
(1)
from
derived
be
can
(3)
and
(2)
{n.b.
)
3
.........(
sinh
3
cosh
2
cosh
2
2
2
,
sinh
3
2
3
2
2
sinh
2
,
2
3
cosh
3
are
here
s
coordinate
parametric
The
:
ation
differenti
parametric
By
2b.
)
2
.........(
tan
3
sec
2
tan
sec
3
sec
2
sec
2
,
tan
sec
3
tan
2
,
sec
3
are
here
s
coordinate
parametric
The
:
ation
differenti
parametric
By
2a.
)
1
(
..........
9
4
0
9
4
0
4
2
9
2
1
4
9
:
ation
differenti
implicit
By
1.
.
1
4
9
hyperbola
he
Consider t
ellipse.
the
of
that
similar to
are
methods
This
2
2
2
2
2
2
t
t
dx
dy
t
e
e
dt
dy
t
e
e
dt
dx
e
e
t
y
e
e
t
x
t
t
t
t
t
dx
dy
t
dt
dy
t
t
dt
dx
t
y
t
x
y
x
dx
dy
dx
dy
y
x
dx
dy
y
x
y
x
y
x
x
x
x
x
x
x
x
x










 









 









 









 


























The gradient function of the hyperbola: generalised.
)
3
.........(
sinh
cosh
sinh
,
cosh
s
coordinate
parametric
With
)
2
.........(
tan
sec
tan
,
sec
s
coordinate
parametric
With
)
1
(
..........
:
that
see
to
difficult
not
is
it
1
hyperbola
the
general
For the
2
2
2
2
2
2
t
a
t
b
dx
dy
t
b
y
t
a
x
t
a
t
b
dx
dy
t
b
y
t
a
x
y
a
x
b
dx
dy
b
y
a
x









The equations of the tangent and normal of the hyperbola.
(2)
into
sub
now
;
2
cosh
;
1
sinh
)
2
,
2
(3
)
sinh
2
,
cosh
3
(
in
find
Or
(1)
into
sub
now
;
)
2
,
2
(3
)
tan
2
,
sec
3
(
in
find
Then
tion
differenta
parametric
using
gradient
the
find
to
required
is
it
If
.
n.b
)
2
.......(
0
cosh
2
6
sinh
6
2
2
3
cosh
2
6
2
2
sinh
6
3
)
cosh
3
(
2
2
)
sinh
2
(
3
)
cosh
3
(
3
2
2
)
sinh
2
(
:
is
tangent
the
of
equation
An
)
sinh
2
,
cosh
3
(
are
s
coordinate
parametric
the
If
3.
)
1
......(
0
sec
2
6
tan
6
2
2
3
sec
2
6
2
2
tan
6
3
)
sec
3
(
2
2
)
tan
2
(
3
)
sec
3
(
3
2
2
)
tan
2
(
:
is
tangent
the
of
equation
An
)
tan
2
,
sec
3
(
are
s
coordinate
parametric
the
If
2.
...(*)
0
6
2
2
3
12
2
2
6
3
)
2
3
(
2
2
)
2
(
3
)
2
3
(
3
2
2
)
2
(
:
is
tangent
the
of
equation
an
So
3
2
2
2
9
2
3
4
9
4
tangent
the
of
gradient
The
:
form
Cartesian
In
1.
:
ways
3
in
determined
be
can
course,
of
),
2
,
2
(3
at
hyperbola
this
o
tangent t
The
.
1
4
9
hyperbola
on the
)
2
,
2
(3
point
he
Consider t
4
2
2






























































t
t
t
t
t
t
t
t
t
t
t
x
y
t
x
t
y
t
x
t
y
t
x
t
y
t
t
t
t
x
y
t
x
t
y
t
x
t
y
t
x
t
y
t
t
x
y
x
y
x
y
x
y
y
x
dx
dy
y
x

v
v
v
v
v
Example:A locus problem involving tangents
)
2
......(
0
cos
sin
0
)
cos
(sin
cos
sin
)
cos
(
sin
cos
)
sin
(
is
of
equation
sin
cos
0
2
2
ation
differenti
implicit
,
circle
the
For
)
1
......(
0
cos
sin
0
)
cos
(sin
cos
sin
)
cos
(
sin
cos
)
sin
(
is
of
equation
an
So
sin
cos
sin
cos
0
2
2
ation
differenti
implicit
1,
ellipse
the
For
.
and
of
equations
parametric
the
find
to
need
we
Here
:
Solution
varies.
as
of
locus
the
Find
.
at
meet
and
).
sin
,
cos
(
point
at the
tangent
has
circle
the
and
)
sin
,
cos
(
point
at the
tangent
has
1
ellipse
The
2
2
2
2
2
2
2
2
1
2
2
2
2
2
2
2
2
2
2
2
1
2
1
2
2
2
2
1
2
2
2
2





















































a
t
x
t
y
t
t
a
t
x
t
y
t
a
x
t
t
t
a
y
l
t
t
y
x
dx
dy
dx
dy
y
x
a
y
x
ab
t
bx
t
ay
t
t
ab
t
bx
t
ay
t
a
x
t
a
t
b
t
b
y
l
t
a
t
b
t
b
a
t
a
b
y
a
x
b
dx
dy
dx
dy
b
y
a
x
b
y
a
x
l
l
t
P
P
l
l
t
a
t
a
l
a
y
x
t
b
t
a
l
b
y
a
x
Contd >>
Example:Alocus problem involving tangents (contd)
 
0)
(
axis
just the
is
varies
as
of
locus
the
)
0
,
cos
(
s
coordinate
has
0
0
sin
)
(
)
2
......(
0
cos
sin
)
1
......(
0
cos
sin
assume
must
we
ellipse
for the
cos
)
(
cos
)
(
cos
)
(
)
2
......(
0
cos
sin
)
1
......(
0
cos
sin
:
(2)
and
(1)
solve
we
find
To
)
2
......(
0
cos
sin
is
of
equation
The
)
1
......(
0
cos
sin
is
of
equation
The
:
Solution
varies.
as
of
locus
the
Find
.
at
meet
and
).
sin
,
cos
(
point
at the
tangent
has
circle
the
and
)
sin
,
cos
(
point
at the
tangent
has
1
ellipse
The
2
2
2
1
2
1
2
2
2
2
1
2
2
2
2











































y
x
t
P
t
a
P
y
t
y
b
a
b
ab
t
bx
t
by
ab
t
bx
t
ay
b
a
t
a
x
a
b
a
t
x
a
b
a
ab
t
x
a
b
a
a
t
ax
t
ay
ab
t
bx
t
ay
P
a
t
x
t
y
l
ab
t
bx
t
ay
l
t
P
P
l
l
t
a
t
a
l
a
y
x
t
b
t
a
l
b
y
a
x
Exercises: Tangents and normals of the ellipse and hyperbola.
)
tan
3
,
sec
5
(
at
1
9
25
2.
)
sin
,
cos
2
(
at
1
4
1.
:
specified
point
at the
sections
conic
following
the
of
normal
and
tangent
the
of
equations
the
Find
2
2
2
2








y
x
y
x
Exercises: Tangents and normals of the ellipse and hyperbola
(solutions).
0
cos
sin
3
sin
2
cos
cos
sin
4
sin
2
cos
sin
cos
)
cos
2
(
sin
2
)
sin
(
cos
)
cos
2
(
cos
sin
2
)
sin
(
:
is
normal
the
of
equation
The
0
2
cos
sin
2
0
)
cos
(sin
2
cos
sin
2
cos
2
cos
sin
2
sin
2
)
cos
2
(
cos
)
sin
(
sin
2
)
cos
2
(
sin
2
cos
)
sin
(
:
is
tangent
the
of
equation
an
So
sin
2
cos
sin
4
cos
2
),
sin
,
cos
2
(
At
4
tangent
the
of
gradient
The
0
2
4
2
1
1
4
.
1
2
2
2
2
2
2




























































































x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
dx
dy
y
x
dx
dy
dx
dy
y
x
y
x
Exercises: Tangents and normals of the ellipse and hyperbola
(solutions).
0
tan
sec
34
tan
5
sec
3
tan
sec
25
tan
5
tan
sec
9
sec
3
)
sec
5
(
tan
5
)
tan
3
(
sec
3
)
sec
5
(
sec
3
tan
5
)
tan
3
(
:
is
normal
the
of
equation
An
0
15
sec
3
tan
5
0
)
tan
(sec
15
sec
3
tan
5
sec
15
sec
3
tan
15
tan
5
)
sec
5
(
sec
3
)
tan
3
(
tan
5
)
sec
5
(
tan
5
sec
3
)
tan
3
(
:
is
tangent
the
of
equation
an
So
tan
5
sec
3
tan
3
25
sec
5
9
),
tan
3
,
sec
5
(
At
25
9
tangent
the
of
gradient
The
0
9
2
25
2
1
9
25
.
2
2
2
2
2
2
2



























































































x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
dx
dy
y
x
dx
dy
dx
dy
y
x
y
x
•Section 5. Typical examination
questions.
Example 1

2
0
for
,
of
values
possible
the
determine
b)
origin,
the
is
where
,
3
that
and
,
2
3
is
of
ty
eccentrici
Given that
.
of
focus
a
is
and
point
at the
axis
-
x
the
cuts
at
to
normal
The
)
6
(
tan
)
(
sin
is
)
tan
,
sec
(
at
normal
the
of
equation
that the
Show
)
.
1
equation
has
hyperbola
The
2
2
2
2
2
2








t
t
O
OS
OA
C
C
S
A
P
C
t
b
a
by
t
ax
t
b
t
a
P
a
b
y
a
x
C
Contd >>
Example 1

2
0
for
,
of
values
possible
the
determine
b)
origin,
the
is
where
,
3
that
and
,
2
3
is
of
ty
eccentrici
Given that
.
of
focus
a
is
and
point
at the
axis
-
x
the
cuts
at
to
normal
The
)
6
(
tan
)
(
sin
is
)
tan
,
sec
(
at
normal
the
of
equation
that the
Show
)
.
1
equation
has
hyperbola
The
2
2
2
2
2
2








t
t
O
OS
OA
C
C
S
A
P
C
t
b
a
by
t
ax
t
b
t
a
P
a
b
y
a
x
C
Example 2
Example 3
Contd >>
Example 3
●
S (5, 0)
P (5, 9/4) ●
● R
Example 4
Contd >>
v
v
v
vvvv
vvvv vvvv vvvv
vvvvv
vvvvv
vvvvv
Example 4 contd.
v v v
v
v
v
v
v
v v
v

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Mastering the Unannounced Regulatory Inspection
 

Further pure mathematics 3 coordinate systems

  • 2. Philosophy of the course • The course does not aim to present a teaching plan for any of the FP3 topics. This is left to your professional judgement as a teacher. • The course is more about presenting each topic in a way that provides insights into the topic beyond the syllabus. • In general the course will present the necessary knowledge and skills to teach the Edexcel FP3 module effectively. • Relevant examination problem solving techniques will be demonstrated.
  • 3. Session structure • Section 1: Introduction to the ellipse and hyperbola. • Section 2: Cartesian and parametric equations of the ellipse and hyperbola. • Section 3: Properties of the ellipse and hyperbola: eccentricity, foci and directrices. • Section 4: Tangents and normals to the ellipse and hyperbola. • Section 5. Typical examination questions.
  • 4. •Section 1: Introduction to the ellipse and hyperbola.
  • 5. We get the hyperbola and the ellipse. If a double cone is cut in a certain way. The circle, ellipse, parabola and hyperbola are called the conic sections. This web page http://bit.ly/1dJzJ1Y gives an animation of how an ellipse might be generated from a cone. Essentially this topic extends the study of FP1 Coordinate geometry which introduced some fundamental properties of the parabola and hyperbola. Here we study the ellipse and the hyperbola to a greater depth. The ellipse and the hyperbola are from a family of curves called the conic sections.
  • 6. • Apollonius of Perga (approx. 262 BC–190 BC) was a Greek geometer who studied with Euclid. He is best known for his work on cross sections of a cone. This is the first recorded study of the conic sections and led to a school of mathematics which developed the important work of Apollonius. • Important Arabic additions to the conics from the tenth and eleventh century were also crucial to the mathematics of the scientific revolution. • In fact, even to this day, large parts The Conics of Apollonius do not exist in their original Greek form, and are known to us only through Arabic translations. • http://www.quadrivium.info/MathInt/Notes/Apollonius.pdf Abrief history of the conic sections.
  • 7. The importance of the ellipse. Kepler model of the elliptical orbits of planets in a solar system → The ellipse was brought into scientific prominence when Johannes Kepler proved that planetary orbits were not circular but elliptical. http://csep10.phys.utk.edu/astr161/lect/history/kepler.html Keplers laws of planetary motion: First Law: The orbit of every planet is an ellipse with the sun at one of the foci. Second Law: A line joining a planet and the Sun sweeps out equal areas during equal intervals of time. Third Law: The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
  • 8. Hyperbolae are also important in astronomy: The majority of comets are in hyperbolic or parabolic orbits. If a comet’s initial velocity is insignificant the path is parabolic, whereas if the initial velocity is significant, the path is hyperbolic. http://bit.ly/1f5Xi21 Ellipses and hyperbolae are also important in other aspects of science and in economics. You can find more here: http://bit.ly/1igiw26
  • 9. •Section 2: Cartesian and parametric equations of the ellipse and hyperbola.
  • 10. The ellipse as a transformed circle. a O . , value fixed by the circle on the points all of coordinate he Multiply t : way in this circle the transform Now a b a b P y  : 0 , circle he Consider t 2 2 2    a a y x ●(x, y) ●       y a b x, circle. on the lies ) , ( then ellipse on the lies ) , ( If : follows as found be can ellipse the of equation The Y b a X Y X ellipse. general the of equation Cartesian The 1 So 2 2 2 2 2 2 2            b Y a X a Y b a X red. in shown ellipse, an is circle ed transform The
  • 11. Fundamental properties of the ellipse. A: (a, 0) O b y b y x y a x a x y x b y a x               1 0 when axis the intersects 1 0 when axis the intersects 1 equation Cartesian with ellipse The 2 2 2 2 2 2 2 2 B (–a, 0) C (0, b) D (0, –b) If a > b then AB is called the major axis and CD is called the minor axis of the ellipse. If the other case when b < a, AB will be the minor and CD the major axis.
  • 12. Parametric equations of the ellipse. (a, 0) O diagram. in the indicated are 2π , 2 3π π, , 2 π , 0 with ellipse on the points The (1). equation Cartesian with ellipse this of equations parametric the gives ) 2 ( ) 2 ( .......... sin , cos or sin , cos writing to us leads This 1 sin cos it with comparing by ed parametris be can ) 1 .( .......... 1 equation Cartesian with ellipse The 2 2 2 2 2 2          t t b y t a x t b y t a x t t b y a x (0, b) t = 0 t = 𝟏 𝟐  t =  t = 3 2  t = 2
  • 13. Rectangular to general hyperbolae                                                                             hyperbola ed transform the of equation the : 2 2 are s coordinate new the So matrix the by ed transform are s coordinate the know we FP1 From clockwise. 4 by graph this rotate we Now graph has This FP1. in met hyperbola r rectangula he Consider t 2 2 2 2 2 2 1 2 2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 c Y X xy Y X y x y x Y X y x y x Y X y x Y X c xy  ↓ rotate ¼ clockwise Note: x = 0 and y = 0 are lines of symmetry
  • 14. Rectangular to general hyperbolae      c Y X c xy 2 2 2 Perpendicular asymptotes. Perpendicular asymptotes. later. verified be will This : 1 is 0 and 0 symmetry of lines with hyperbolae general the of equation Cartesian The asymptotes lar perpendicu have not do hyperbolae general In 2 2 2 2      b y a x y x
  • 15. Properties and parametric equations of the hyperbola. ) 1 ....( tan , sec s coordinate parametric implies 1 tan sec identity The ). 0 , ( and ) 0 , ( at axis Intersects 1 2 2 2 2 2 2 t b y t a x t t a a x b y a x         (– a, 0) (a, 0) t = 0 t =  ) 2 ....( sinh , cosh use we , For . 1 cosh because when applies only But this ) 2 ....( sinh , cosh s coordinate parametric implies 1 sinh cosh and 1 Also 2 2 2 2 2 2 a t b y t a x a x t a x t b y t a x t t b y a x             (a, 0) t = 0
  • 16. Exercises: Equations of the ellipse and hyperbola functions. tric trigonome of in terms form parametric the Give . graph. sketch a Provide . 1 form equivalent the Give : above equations the of each For 144 16 36 . 4 36 9 4 . 3 36 4 9 . 2 16 4 . 1 2 2 2 2 2 2 2 2 2 2 2 2 iii ii b y a x i. y x y x y x y x          
  • 17. Exercises: Equations of the ellipse and hyperbola (solutions) . sin 2 , cos 4 : is form parametric The . : is graph sketch A . 1 2 4 16 4 16 4 . 1 2 2 2 2 2 2 2 2            y x iii ii y x y x i. y x 4 – 4 – 2 2 . sin 3 , cos 2 : is form parametric The . : is graph sketch A . 1 3 2 36 4 9 36 4 9 . 2 2 2 2 2 2 2 2 2            y x iii ii y x y x i. y x 2 – 2 – 3 3 . tan 2 , sec 3 : is form parametric The . : is graph sketch A . 1 2 3 36 9 4 36 9 4 . 3 2 2 2 2 2 2 2 2            y x iii ii y x y x i. y x 3 – 3 . tan 3 , sec 2 : is form parametric The . : is graph sketch A . 1 3 2 144 16 36 144 16 36 . 4 2 2 2 2 2 2 2 2            y x iii ii y x y x i. y x 2 – 2
  • 18. •Section 3: Properties of the ellipse and hyperbola: eccentricity, foci and directrices.
  • 19. Geometrical derivation of the ellipse. •One way (but not the only one) to define the conic sections - parabola, hyperbola and ellipse – is the following : •They are the loci of points P in the plane that satisfy the following condition: •The distance of P from a fixed point S = e the distance of P from a fixed line L. •The fixed point S is called the focus of the conic section. •The fixed line L is called its directrix. •The constant multiple e is called its eccentricity. • e =1 for the parabola, 0 < e < 1 for the ellipse and e > 1 for the hyperbola.
  • 20. Focus, directrix and Cartesian equation of the parabola. ● S (a, 0) L: x = – a Let the focus be the point S (a, 0) and the directrix L: x = – a. Points P (x, y) on the parabola satisfy: PS = 1PT. ● P (x, y) T (– a , y)● Or PS2 = PT2 ax y ax a x y ax a x a x y a x 4 2 2 ) ( ) ( 2 2 2 2 2 2 2 2 2              
  • 21. Focus, directrix and Cartesian equation of the ellipse. ● S (ae, 0) L: x = a/e Here we let the focus be the point S (ae, 0) and the directrix L: x = a/e, where 0 < e < 1, is the eccentricity of the ellipse. Points P (x, y) on the ellipse satisfy: PS = ePT : 0 < e < 1 ● P (x, y) ● T (a/e , y) Or PS2 = e2PT2 ) 1 ( where , 1 or 1 ) 1 ( ) 1 ( ) 1 ( 2 2 ) ( ) ( 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 e a b b y a x e a y a x e a y e x aex a x e y aex e a x x e a e y ae x                         
  • 22. Focus, directrix and Cartesian equation of the ellipse. ● S (ae, 0) L: x = a/e As the equation of the ellipse has even powers of x and of y it is symmetrical about both axes. This symmetry implies that the ellipse has two foci and two directrices. The other focus is S′ (–ae, 0) and other directrix is L′ : x = –a/e. ● P (x, y) ● T (a/e , y) . 1 0 ), 1 ( where , 1 is ellipse the of equation general The 2 2 2 2 2 2 2       e e a b b y a x L′ : x = –a/e S′ (–ae, 0) ● ) 0 , ( and ) 0 , ( are intercepts the So . 1 0 ) 0 , ( and ) 0 , ( are intercepts the So . 1 0 2 2 2 2 b b y b y b y x a a x a x a x y               (a, 0) (–a, 0) (–b, 0) (b, 0)
  • 23. Focus, directrix and Cartesian equation of the hyperbola. ● S (ae, 0) L: x = a/e Here we let the focus be the point S (ae, 0) and the directrix L: x = a/e, where e, e >1, is the eccentricity of the hyperbola. Points P (x, y) on the hyperbola satisfy: PS = ePT : e > 1 ● P (x, y) T (a/e , y)● Or PS2 = e2PT2 0 ensure to : ) 1 ( write to need t we reason tha the Note ) 1 ( where , 1 or 1 ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( 2 2 ) ( ) ( 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2                                 b e e a b b y a x e a y a x e a y e x e a y e x aex a x e y aex e a x e a x e y ae x
  • 24. Focus, directrix and Cartesian equation of the hyperbola. Again the equation of the hyperbola has even powers of x and of y it is symmetrical about both axes. The symmetry implies that it has two foci and two directrices. The other focus is S′ (–ae, 0) and other directrix is L′ : x = –a/e. . 1 ), 1 ( where , 1 is hyperbola the of equation general The 2 2 2 2 2 2 2      e e a b b y a x . intercepts no Hence . impossible is which , 1 0 ) 0 , ( and ) 0 , ( are intercepts the So . 1 0 2 2 2 2 y b y x a a x a x a x y            (a, 0)● ● S (ae, 0) L: x = a/e T (a/e , y)● L′ : x = –a/e S′ (–ae, 0) ● ● P (x, y) (–a, 0) ●
  • 25. Example: Focus and directrix of the ellipse. 12 12 ) 4 1 1 ( 16 ), 1 ( As 4 (1) Then ) ellipse the for 1 0 ( 2 1 4 1 2 8 ) 2 ( and ) 1 ( ) 2 ( .......... 2 So ). 0 , ( 0) (2, focus The ) 1 ( .......... 8 So . 8 directrix The : 2 2 2 2 2                       b b e a b a e e e e e e a ae e a e a x Solution
  • 26. Example: Focus and directrix of the hyperbola. 5 16 are s directrice The ) 0 , 5 ( ) 0 , ( are foci The hyperbola) for the 1 ( 4 5 16 25 16 9 ) 1 ( ) 1 ( 16 9 So ) 1 ( hyperbola For the . 3 and 4 get we 1 form standard with the 1 9 16 Comparing . directices and foci the showing hyperbola Sketch the s. directrice its of equations the and foci its of s coordinate the Find 1 9 16 equation has hyperbola A 2 2 2 2 2 2 2 2 2 2 2 2 2 2                            e a x ae e e e e e e a b b a b y a x y x Solution: y x ● S (5, 0) S ′(– 5, 0) ● x = 3.2 x = – 3.2
  • 27. Exercises: Focus and directrix above. sections conic the of each for s directrice and foci the Find 1 7 16 . 3 1 3 4 . 2 1 3 4 . 1 2 2 2 2 2 2       y x y x y x
  • 28. Exercises: Focus and directrix (solutions) 3 16 s Directrice ); 0 , 3 ( ) 0 , ( Foci . 4 3 . 16 9 ) 1 ( 16 7 ) 1 ( . 7 , 16 ellipse. An . 1 7 16 . 3 7 7 4 7 4 s Directrice ); 0 , 7 ( ) 0 , ( Foci . 2 7 . 4 7 ) 1 ( 4 3 so ), 1 ( Here . 3 , 4 1 : hyperbola a of equation the is This . 1 3 4 . 2 4 s Directrice ); 0 , 1 ( ) 0 , ( Foci . 2 1 . 4 1 ) 1 ( 4 3 so ), 1 ( Here . 3 , 4 1 : ellipse an of equation the is This . 1 3 4 . 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2                                                                   e a x ae e e e e a b b a y x e a x ae e e e e a b b a b y a x y x e a x ae e e e e a b b a b y a x y x
  • 29. •Section 4: Tangents and normals to the ellipse and hyperbola.
  • 30. The gradient function of the ellipse. (1)} in s coordinate parametric the subs by (1) from derived be can (2) {n.b. ) 2 .........( sin 3 cos 2 cos 2 , sin 2 sin 2 , cos 3 are 1 4 9 of s coordinate parametric The : ation differenti parametric By 2. ) 1 ( .......... 9 4 0 9 4 0 4 2 9 2 1 4 9 : ation differenti implicit By 1. : ways in two determined be can function gradient The . 1 4 9 ellipse he Consider t 2 2 2 2 2 2 t t dx dy dt dx dt dy dx dy t dt dy t dt dx t y t x y x y x dx dy dx dy y x dx dy y x y x y x                          
  • 31. The gradient function of the ellipse generalised. ) 2 .........( sin cos ) 1 ( .......... . 1 ellipse For the : general made be easily can slide previous the of n calculatio The 2 2 2 2 2 2 t a t b dx dy y a x b dx dy b y a x      
  • 32. The equations of the tangent and normal of the ellipse. before) as ( 3 2 sin 3 cos 2 sin 3 cos 2 tangent the of gradient Then the n calculatio or inspection By ). , ( ) sin 2 , cos 3 ( in find to need tion we differenta parametric using gradient the find To . n.b 0 cos 6 sin 6 2 3 or cos 6 2 sin 6 3 ) cos 3 ( 2 ) sin 2 ( 3 ) cos 3 ( 3 2 ) sin 2 ( : is tangent the of equation an So ) sin 2 , cos 3 ( are s coordinate parametric The : form parametric In 2. 0 2 6 2 3 2 3 ) ( 2 ) ( 3 ) ( 3 2 ) ( : is tangent the of equation an So 3 2 9 4 9 4 tangent the of gradient The : form Cartesian In 1. : ways in two determined be can ) , ( at ellipse this o tangent t the of equation An . 1 4 9 ellipse on the ) , ( point he Consider t 4 4 4 2 2 2 2 2 3 2 2 6 2 2 6 2 2 3 2 2 2 2 2 3 2 2 2 2 2 2 2 2 3 2 2 2 2 2 3 2 2 2 2 2 2 2 3                                                            t t dx dy t t t t t t x y t x t y t x t y t x t y t t x y x y x y x y y x dx dy y x
  • 33. The gradient function of the hyperbola (1)} in s coordinate parametric relevant the subs by (1) from derived be can (3) and (2) {n.b. ) 3 .........( sinh 3 cosh 2 cosh 2 2 2 , sinh 3 2 3 2 2 sinh 2 , 2 3 cosh 3 are here s coordinate parametric The : ation differenti parametric By 2b. ) 2 .........( tan 3 sec 2 tan sec 3 sec 2 sec 2 , tan sec 3 tan 2 , sec 3 are here s coordinate parametric The : ation differenti parametric By 2a. ) 1 ( .......... 9 4 0 9 4 0 4 2 9 2 1 4 9 : ation differenti implicit By 1. . 1 4 9 hyperbola he Consider t ellipse. the of that similar to are methods This 2 2 2 2 2 2 t t dx dy t e e dt dy t e e dt dx e e t y e e t x t t t t t dx dy t dt dy t t dt dx t y t x y x dx dy dx dy y x dx dy y x y x y x x x x x x x x x                                                                       
  • 34. The gradient function of the hyperbola: generalised. ) 3 .........( sinh cosh sinh , cosh s coordinate parametric With ) 2 .........( tan sec tan , sec s coordinate parametric With ) 1 ( .......... : that see to difficult not is it 1 hyperbola the general For the 2 2 2 2 2 2 t a t b dx dy t b y t a x t a t b dx dy t b y t a x y a x b dx dy b y a x         
  • 35. The equations of the tangent and normal of the hyperbola. (2) into sub now ; 2 cosh ; 1 sinh ) 2 , 2 (3 ) sinh 2 , cosh 3 ( in find Or (1) into sub now ; ) 2 , 2 (3 ) tan 2 , sec 3 ( in find Then tion differenta parametric using gradient the find to required is it If . n.b ) 2 .......( 0 cosh 2 6 sinh 6 2 2 3 cosh 2 6 2 2 sinh 6 3 ) cosh 3 ( 2 2 ) sinh 2 ( 3 ) cosh 3 ( 3 2 2 ) sinh 2 ( : is tangent the of equation An ) sinh 2 , cosh 3 ( are s coordinate parametric the If 3. ) 1 ......( 0 sec 2 6 tan 6 2 2 3 sec 2 6 2 2 tan 6 3 ) sec 3 ( 2 2 ) tan 2 ( 3 ) sec 3 ( 3 2 2 ) tan 2 ( : is tangent the of equation An ) tan 2 , sec 3 ( are s coordinate parametric the If 2. ...(*) 0 6 2 2 3 12 2 2 6 3 ) 2 3 ( 2 2 ) 2 ( 3 ) 2 3 ( 3 2 2 ) 2 ( : is tangent the of equation an So 3 2 2 2 9 2 3 4 9 4 tangent the of gradient The : form Cartesian In 1. : ways 3 in determined be can course, of ), 2 , 2 (3 at hyperbola this o tangent t The . 1 4 9 hyperbola on the ) 2 , 2 (3 point he Consider t 4 2 2                                                               t t t t t t t t t t t x y t x t y t x t y t x t y t t t t x y t x t y t x t y t x t y t t x y x y x y x y y x dx dy y x  v v v v v
  • 36. Example:A locus problem involving tangents ) 2 ......( 0 cos sin 0 ) cos (sin cos sin ) cos ( sin cos ) sin ( is of equation sin cos 0 2 2 ation differenti implicit , circle the For ) 1 ......( 0 cos sin 0 ) cos (sin cos sin ) cos ( sin cos ) sin ( is of equation an So sin cos sin cos 0 2 2 ation differenti implicit 1, ellipse the For . and of equations parametric the find to need we Here : Solution varies. as of locus the Find . at meet and ). sin , cos ( point at the tangent has circle the and ) sin , cos ( point at the tangent has 1 ellipse The 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 1 2 2 2 2                                                      a t x t y t t a t x t y t a x t t t a y l t t y x dx dy dx dy y x a y x ab t bx t ay t t ab t bx t ay t a x t a t b t b y l t a t b t b a t a b y a x b dx dy dx dy b y a x b y a x l l t P P l l t a t a l a y x t b t a l b y a x Contd >>
  • 37. Example:Alocus problem involving tangents (contd)   0) ( axis just the is varies as of locus the ) 0 , cos ( s coordinate has 0 0 sin ) ( ) 2 ......( 0 cos sin ) 1 ......( 0 cos sin assume must we ellipse for the cos ) ( cos ) ( cos ) ( ) 2 ......( 0 cos sin ) 1 ......( 0 cos sin : (2) and (1) solve we find To ) 2 ......( 0 cos sin is of equation The ) 1 ......( 0 cos sin is of equation The : Solution varies. as of locus the Find . at meet and ). sin , cos ( point at the tangent has circle the and ) sin , cos ( point at the tangent has 1 ellipse The 2 2 2 1 2 1 2 2 2 2 1 2 2 2 2                                            y x t P t a P y t y b a b ab t bx t by ab t bx t ay b a t a x a b a t x a b a ab t x a b a a t ax t ay ab t bx t ay P a t x t y l ab t bx t ay l t P P l l t a t a l a y x t b t a l b y a x
  • 38. Exercises: Tangents and normals of the ellipse and hyperbola. ) tan 3 , sec 5 ( at 1 9 25 2. ) sin , cos 2 ( at 1 4 1. : specified point at the sections conic following the of normal and tangent the of equations the Find 2 2 2 2         y x y x
  • 39. Exercises: Tangents and normals of the ellipse and hyperbola (solutions). 0 cos sin 3 sin 2 cos cos sin 4 sin 2 cos sin cos ) cos 2 ( sin 2 ) sin ( cos ) cos 2 ( cos sin 2 ) sin ( : is normal the of equation The 0 2 cos sin 2 0 ) cos (sin 2 cos sin 2 cos 2 cos sin 2 sin 2 ) cos 2 ( cos ) sin ( sin 2 ) cos 2 ( sin 2 cos ) sin ( : is tangent the of equation an So sin 2 cos sin 4 cos 2 ), sin , cos 2 ( At 4 tangent the of gradient The 0 2 4 2 1 1 4 . 1 2 2 2 2 2 2                                                                                             x y x y x y x y x y x y x y x y x y dx dy y x dx dy dx dy y x y x
  • 40. Exercises: Tangents and normals of the ellipse and hyperbola (solutions). 0 tan sec 34 tan 5 sec 3 tan sec 25 tan 5 tan sec 9 sec 3 ) sec 5 ( tan 5 ) tan 3 ( sec 3 ) sec 5 ( sec 3 tan 5 ) tan 3 ( : is normal the of equation An 0 15 sec 3 tan 5 0 ) tan (sec 15 sec 3 tan 5 sec 15 sec 3 tan 15 tan 5 ) sec 5 ( sec 3 ) tan 3 ( tan 5 ) sec 5 ( tan 5 sec 3 ) tan 3 ( : is tangent the of equation an So tan 5 sec 3 tan 3 25 sec 5 9 ), tan 3 , sec 5 ( At 25 9 tangent the of gradient The 0 9 2 25 2 1 9 25 . 2 2 2 2 2 2 2                                                                                            x y x y x y x y x y x y x y x y x y dx dy y x dx dy dx dy y x y x
  • 41. •Section 5. Typical examination questions.
  • 42. Example 1  2 0 for , of values possible the determine b) origin, the is where , 3 that and , 2 3 is of ty eccentrici Given that . of focus a is and point at the axis - x the cuts at to normal The ) 6 ( tan ) ( sin is ) tan , sec ( at normal the of equation that the Show ) . 1 equation has hyperbola The 2 2 2 2 2 2         t t O OS OA C C S A P C t b a by t ax t b t a P a b y a x C Contd >>
  • 43. Example 1  2 0 for , of values possible the determine b) origin, the is where , 3 that and , 2 3 is of ty eccentrici Given that . of focus a is and point at the axis - x the cuts at to normal The ) 6 ( tan ) ( sin is ) tan , sec ( at normal the of equation that the Show ) . 1 equation has hyperbola The 2 2 2 2 2 2         t t O OS OA C C S A P C t b a by t ax t b t a P a b y a x C
  • 46. Example 3 ● S (5, 0) P (5, 9/4) ● ● R
  • 47. Example 4 Contd >> v v v vvvv vvvv vvvv vvvv vvvvv vvvvv vvvvv
  • 48. Example 4 contd. v v v v v v v v v v v