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.
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
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 = 1PT.
● 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 = ePT : 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 = ePT : 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
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