Hyperbola P a/e e>1 O: center F1, F2: foci V1, V2: vertices PF2 – PF1 = V1F2 – V1F1 = V1V2 = 2a...
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Hyperbola 1 2 2 2 2 b y a x P a/e e>1 O: center F1, F2: foci V1, V2: vertices PF2 – PF1 = V1F2 – V1F1 = V1V2 = 2a Product of shortest distances from P to the asymptotes is a constant. F2 F1 Directri x a ae Asymptot e O V2 V1 b When the asymptotes are perpendicular it is a called a rectangular hyperbola. Hyperbol a Axis
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Transcript of Hyperbola P a/e e>1 O: center F1, F2: foci V1, V2: vertices PF2 – PF1 = V1F2 – V1F1 = V1V2 = 2a...
Hyperbola
12
2
2
2
b
y
a
x
P
a/e
e>1O: centerF1, F2: fociV1, V2: verticesPF2 – PF1 = V1F2 – V1F1 = V1V2 = 2aProduct of shortest distances from P to the asymptotes is a constant.
F2 F1
Directrix
a
aeAsymptote
OV2 V1
b
When the asymptotes are perpendicular it is a called a rectangular hyperbola.
Hyperbola
Axis
2
HYPERBOLA DIRECTRIX-FOCUS METHOD
Draw an ellipse, focus is 50 mm from the directrix and the eccentricity is 3/2
F1 ( focus)
DI R
EC
TR
IX
V(vertex)
A
B
E
C
1’
1
P1
P1’
P2
P2’
VE = VF1
F1-P1=F1-P1’ = 1-1’
2
2’
F1-P2=F1-P2’= 2-2’P2 AND P2’ ALSO LIE ON THE HYPERBOLA
F1-P1/(P1 to directrix AB) =1-1’/C-1=VE/VC (similar triangles)=VF1/VC=2/3THEREFORE P1 AND P1’ LIE ON THE HYPERBOLA
P
O
40 mm
30 mm
1’
2’
3’
4’5’ 1 2 3
4
5
HYPERBOLATHROUGH A POINT
OF KNOWN CO-ORDINATESSolution Steps:1) Extend horizontal line from P to right side. 2) Extend vertical line from P upward.3) On horizontal line from P, mark some points taking any distance and name them 1, 2, 3 etc.4) Join 1-2-3 points to pole O. Let them cut part [P-B] at 1’,2’,3’ points.5) From horizontal 1,2,3 draw vertical lines downwards and6) From vertical 1’,2’,3’ points [from P-B] draw horizontal lines.7) Vertical line from 1 and horizontal line from 1’ P1.Similarly mark P2,
P3, P4 points.
8) Repeat the procedure by marking points 4, 5 on upward vertical line from P and joining all those to pole O. They cut the horizontal line from P at 4’ and 5’. Repeat earlier procedure to obtain points P4, P5.Join them by a smooth curve.
Problem: Point P is 40 mm and 30 mm from horizontal and vertical axes respectively. Draw a Hyperbola through it.
B
P1
P2
P3
P5
P4
Hyperbola-rectangle method
1
2
1’ 2’
Axis height
Base
Height of hyperbola
VOLUME:( M3 )
PRE
SSU
RE
( K
g/cm
2 )
0 1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
HYPERBOLAP-V DIAGRAM
Problem: A sample of gas is expanded in a cylinderfrom 10 unit pressure to 1 unit pressure. Expansion follows law PV=Constant. If initial volume being 1 unit, draw the curve of expansion. Also Name the curve.
Form a table giving few more values of P & V
P V = C+
10542.521
122.54510
101010101010
++
++
++
======
Now draw a Graph of Pressure against Volume.
It is a PV Diagram and it is a Hyperbola.Take pressure on vertical axis and
Volume on horizontal axis.
D
F1 F21 2 3 4
A B
C
p1
p2
p3
p4
O
Q TANGENT
NO
RM
AL
TO DRAW TANGENT & NORMAL TO THE CURVE AT A GIVEN POINT ( Q )
1. JOIN POINT Q TO F1 & F2
2. BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL3. A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE.
ELLIPSE TANGENT & NORMALProblem :
ELLIPSE TANGENT & NORMAL
F ( focus)
DIR
EC
TR
IX
V
ELLIPSE
(vertex)
A
B
T
T
N
N
Q
900
TO DRAW TANGENT & NORMAL TO THE CURVE
AT A GIVEN POINT ( Q )
1.JOIN POINT Q TO F.2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO ELLIPSE FROM Q5.TO THIS TANGENT DRAW PERPENDICULAR LINE FROM Q. IT IS NORMAL TO CURVE.
Problem:
A
B
PARABOLA
VERTEX F ( focus)
V
Q
T
N
N
T
900
TO DRAW TANGENT & NORMAL TO THE CURVE
AT A GIVEN POINT ( Q )
1.JOIN POINT Q TO F.2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO THE CURVE FROM Q5.TO THIS TANGENT DRAW PERPENDICULAR LINE FROM Q. IT IS NORMAL TO CURVE.
PARABOLATANGENT & NORMALProblem:
F ( focus)V
(vertex)
A
B
HYPERBOLATANGENT & NORMAL
QN
N
T
T
900
TO DRAW TANGENT & NORMAL TO THE CURVE
FROM A GIVEN POINT ( Q )
1.JOIN POINT Q TO F.2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO CURVE FROM Q5.TO THIS TANGENT DRAW PERPENDICULAR LINE FROM Q. IT IS NORMAL TO CURVE.
Problem 16
Concept of Principal lines of a plane
A’
A
C’
B’
TLB
CPoint view
All the points lie on a straight line representing the edge of
the plane
Principal line
T
F
A1T Draw a line on the plane in one view parallel to the other plane.
The corresponding projection in the other plane will give the true length.
Frontal line (parallel to frontal plane)
Principal lines: Lines on the boundary or within the surface, parallel to the principal planes of projection -They can be frontal lines (parallel to frontal plane)-Horizontal lines (parallel to top plane)
a’
b’
c’
a
b
c
True length
T
F
T
F
a’
b’
c’
a
b
c
True length
Horizontal line (parallel to top plane)
f’
l’
f l
l’
l
To obtain the edge view of a plane
T
F
a’
b’
c’
a
b
c
True length
Horizontal line (parallel to top plane)
l’
l
-Draw a principle line in one principle view and project the true length line in the other principle view
-With the reference line perpendicular to the true length line, draw a primary auxiliary view of the plane, to obtain the edge view
Distances:
a1, b1, c1 from x1y1 = a’, b’, c’ from xy respectively
Edge view of the plane
x y
c1
a1
b1
x1
y1
Auxiliary view of TRUE SHAPE of a plane always gives an EDGE VIEW
T
F
a’
b’
c’
a
b
c
True length
Horizontal line (parallel to top plane)
l’
l
x yc2 a2
b2
c1
a1
b1
x1
y1
True shape and dimensions of the plane
True shape is the auxiliary view obtained from the edge view
Edge view of plane
c3 a3
b3
c4
a4
b4Edge view of the plane
is the angle of the plane with the HP
