liceo scientifico “g.ferraris” taranto school year 2011-2012 maths course the hyperbola...
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Liceo Scientifico “G.Ferraris”Taranto
School Year 2011-2012
Maths course The hyperbola
UTILIZZARE SPAZIO PER INSERIRE FOTO/IMMAGINE DI RIFERIMENTO LEZIONE
A hyperbola is an open curve with two branches, the intersection
of a plane with both halves of a double cone. The hyperbola
belongs to a family of curves including parabolas, ellipses, circles.
Conic section
| PF1- PF2 | = const
The hyperbola is the geometric locus of points P which moves so
that, the difference between the distances from P to two fixed
points, called foci, is a constant.
Hyperbola as geometric locus
The equation of the hyperbola can be found by using
the distance formula:
Finding the hyperbola equation
given F1(-c ; 0) and F2(c ; 0) c > 0,
let P (x,y) such that | PF1- PF2 | = 2a a > 0
(c2-a2)x2-a2y2=a2(c2-a2) c > a
From this relation, after eliminating radicals and simplifying,
we obtain the hyperbola equation centred at the origin:
If the x-term is positive, it means that the hyperbola is
horizontal or opening East-West
If we place b2=c2–a2 into the previous equation,
we’ll obtain the following:
Hyperbola in canonical form
The hyperbola foci and vertexes
-a a
b
c
c
-c
The x-intercepts of this curve are given by the points – a and a,
that are called vertexes of the hyperbola.
The points of ordinates –b and b are imaginary y-intercepts.
-bc2=a2+b2
Pythaghorean Theorem
The hyperbola axes
- a ac-c
transverse axis
The transverse axis is the segment whose endpoints are the
vertexes of the hyperbola. Its measure is 2a.
The line passing the origin and perpendicular to the transverse
axis is the conjugate axis.
The hyperbola simmetries
It is a symmetry point for this curve. The coordinate axes are symmetry axes.
The centre of the hyperbola is
the midpoint of the transverse
axis that is the origin.
It ‘s also the midpoint of the
segment connecting the foci.
The positive number “b” is called measure of the conjugate
semi-axis.
Hyperbola position
The hyperbola doesn’t have inner points at the band delimited
by the vertical lines x = - a and x = a, then the curve is formed by
2 branches or arms, as shown in the picture.
branche branche
The horizontal lines passing the ordinates – b and b, with the
vertical lines passing the abscissas - a and a, form a rectangle
whose sides measure 2a and 2b.
The diagonal of this rectangle has the same measure of the
focal length, 2c .
The lines that contain the
diagonals of the rectangle
are the asymptotes of
the hyperbola, they are
endless tangent lines.
These asymptotes pass the origin and their equations are of type
y = mx where m = b/a v m = - b/a.
The asymptotes of a hyperbola
a = 4 is the semi-transverse axis
b = 3 is the semi-conjugate axis
c = 5 is the distance from the centre to each focus
y = x are the asymptotes
Example
If a=4 , b=3 and the foci
are horizontally aligned,
the equation is:
43
If the 2 foci are vertically aligned, the x-term is negative and
the equation of the hyperbola becomes:
Vertical hyperbola
12
2
2
2
b
y
a
x
In this case the transverse axis is on the y-axis and its length is 2b.
It means that the hyperbola is opening North-South.
The equations of the asymptotes never change.
We define eccentricity of the hyperbola, the ratio of the focal
length to the measure of the transverse axis.
This ratio is denoted by “e”, that is e = 2c/2b, e = c/b.
Hyperbola eccentricity
This number “e” is always
greater than 1 and defines
the hyperbola opening.
e1 < e2
e1
e2
The more the number “e” is over 1, that is the foci move away
from the vertexes, the more the hyperbola opens.
Eccentricity variation
If a=b, the measure of the conjugate and transverse axes is
the same, then the hyperbola is called equilateral.
Turning this curve 45° around the centre, the asymptotes
coincide with the coordinate axes.
Equilateral hyperbola
equilateral hyperbola
referred to the asymptotesK>0
K<0
xy = k k≠0
Boyle’s law:
PV=k
The case of k>0 represents the law of the inverse proportionality.
Tuscany
Cooling towers of the geysers
KÕbe Port Tower-Japanby Vladimir Shukhov
Cathedral of Brasiliaby Oscar Niemeyer
Course TeacherRosanna Biffi
Linguistic Support Flaviana Ciocia
Performed byTeacher: Rosanna Biffi
Students: Arnesano Alessandro, Basile Giulia, Biondolillo Alessia, Bruno Marianna, D’andria Roberta, Manco Marcello
(Grade 5 D - Secondary High School)
Acknowledgement Marco Dal Bosco
Headmaster
Technical Supporteni
Director Rosanna Biffi
Copyright 2012 © eni S.p.A.