1. Hyperbola Directrix 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.
When the asymptotes are perpendicular it is a called a rectangular hyperbola. P
Axis b F2 V2 O V1 F1
Hyperbola
a/e a
Asymptote ae

2. DIRECTRIX-FOCUS METHOD
Draw an ellipse, focus is 50 mm from the directrix
and the eccentricity is 3/2
HYPERBOLA
DIRECTRIX-FOCUS METHOD 2’ P2 A 1’
VE = VF1 P1 E
DIRECTRIX
F1-P1=F1-P1’ = 1-1’
F1-P1/(P1 to directrix AB) =
1-1’/C-1=VE/VC (similar triangles)
=VF1/VC=2/3
THEREFORE P1 AND P1’ LIE ON THE HYPERBOLA
(vertex) V
F1 ( focus) C 1 2
F1-P2=F1-P2’= 2-2’
P2 AND P2’ ALSO LIE ON THE HYPERBOLA
P1’ B
P2’

3. HYPERBOLA THROUGH A POINT OF KNOWN CO-ORDINATES
Problem: Point P is 40 mm and 30 mm from horizontal
and vertical axes respectively. Draw a Hyperbola through it.
Solution 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 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 and
6)      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. 5 P5 4 P4 5’ 4’ P 1 2 3
40 mm P1 1’ P2 2’ P3 3’
30 mm O B

4. Hyperbola-rectangle method1’ 2’ 2 1
Base
Height of hyperbola
Axis height

5. HYPERBOLA P-V DIAGRAM 1 2 3 4 5 6 7 8 9 10 + PRESSURE ( Kg/cm2) 1 2 3
Problem: A sample of gas is expanded in a cylinder
from 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. 1 2 3 4 5 6 7 8 9 10
Form a table giving few more values of P & V
P V = C + 10 5 4
2.5 2 1 =
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.
PRESSURE
( Kg/cm2) 1 2 3 4 5 6 7 8 9 10
VOLUME:( M3 )

6. TO DRAW TANGENT & NORMAL TO THE CURVE AT A GIVEN POINT ( Q )
ELLIPSE
TANGENT & NORMAL
Problem :
TO DRAW TANGENT & NORMAL
TO THE CURVE AT A GIVEN POINT ( Q )
JOIN POINT Q TO F1 & F2
BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL
A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE. D F1 F2 A B C p1 p2 p3 p4 O
NORMAL Q
TANGENT

7. TO DRAW TANGENT & NORMAL
ELLIPSE
TANGENT & NORMAL
Problem:
TO DRAW TANGENT & NORMAL
TO THE CURVE
AT A GIVEN POINT ( Q )
F ( focus)
DIRECTRIX V
ELLIPSE
(vertex) A B
1.JOIN POINT Q TO F.
2.CONSTRUCT 900 ANGLE WITH
THIS LINE AT POINT F
3.EXTEND THE LINE TO MEET DIRECTRIX
AT T
4. JOIN THIS POINT TO Q AND EXTEND. THIS IS
TANGENT TO ELLIPSE FROM Q
5.TO THIS TANGENT DRAW PERPENDICULAR
LINE FROM Q. IT IS NORMAL TO CURVE. T
900 N Q N T

8. TO DRAW TANGENT & NORMAL
PARABOLA
TANGENT & NORMAL
Problem:
TO DRAW TANGENT & NORMAL
TO THE CURVE
AT A GIVEN POINT ( Q ) A B
PARABOLA
VERTEX F
( focus) V T
1.JOIN POINT Q TO F.
2.CONSTRUCT 900 ANGLE WITH
THIS LINE AT POINT F
3.EXTEND THE LINE TO MEET DIRECTRIX
AT T
4. JOIN THIS POINT TO Q AND EXTEND. THIS IS
TANGENT TO THE CURVE FROM Q
5.TO THIS TANGENT DRAW PERPENDICULAR
LINE FROM Q. IT IS NORMAL TO CURVE.
900 N Q N T

9. TO DRAW TANGENT & NORMAL
HYPERBOLA
TANGENT & NORMAL
Problem 16
TO DRAW TANGENT & NORMAL
TO THE CURVE
FROM A GIVEN POINT ( Q )
F ( focus) V
(vertex) A B
1.JOIN POINT Q TO F.
2.CONSTRUCT 900 ANGLE WITH THIS LINE AT
POINT F
3.EXTEND THE LINE TO MEET DIRECTRIX AT T
4. JOIN THIS POINT TO Q AND EXTEND. THIS IS
TANGENT TO CURVE FROM Q
5.TO THIS TANGENT DRAW PERPENDICULAR
LINE FROM Q. IT IS NORMAL TO CURVE. T
900 N N Q T

10. Concept of Principal lines of a plane
All the points lie on a straight line representing the edge of the plane
Point view C B TL A1 T
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. A T F A’ C’
Principal line B’

11. 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)
Frontal line (parallel to frontal plane) b
True length b l a f l a c T c T F c’ F c’ a’ l’ f’ a’
True length b’ l’
Horizontal line (parallel to top plane) b’

12. To obtain the edge view of a planec1 a1 b1 x1 y1
Edge view of the plane b
True length l a c
-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 T x y c’ F l’ a’
Horizontal line (parallel to top plane) b’

13. Auxiliary view of TRUE SHAPE of a plane always gives an EDGE VIEW
True shape is the auxiliary view obtained from the edge view c1 a1 b1 x1 y1 a c2 a2 b2 c4 a4 b4
Edge view of the plane
a is the angle of the plane with the HP b
True length l c3 a3 b3 a c
Edge view of plane T
True shape and dimensions of the plane y x F c’ l’ a’ b’
Horizontal line (parallel to top plane)