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ENGINEERING CURVES
Part- I {Conic Sections}
ELLIPSE

PARABOLA

HYPERBOLA

1.Concentric Circle Method

1.Rectangle Method

2.Rectangle Method

2 Method of Tangents
( Triangle Method)

1.Rectangular Hyperbola
(coordinates given)

3.Oblong Method
4.Arcs of Circle Method

3.Basic Locus Method
(Directrix – focus)

2 Rectangular Hyperbola
(P-V diagram - Equation given)
3.Basic Locus Method
(Directrix – focus)

5.Rhombus Metho
6.Basic Locus Method
(Directrix – focus)

Methods of Drawing
Tangents & Normals
To These Curves.
ENGINEERING CURVES
CONIC SECTIONS
ELLIPSE, PARABOLA AND HYPERBOLA ARE CALLED CONIC SECTIONS
BECAUSE
THESE CURVES APPEAR ON THE SURFACE OF A CONE
WHEN IT IS CUT BY SOME TYPICAL CUTTING PLANES.
OBSERVE
ILLUSTRATIONS
GIVEN BELOW..

Ellipse

Ξ²

e<1

Ξ±

Ξ± Ξ²

Ξ²< Ξ±

Ξ²> Ξ±

Section Plane
Through all the Generators

Section Plane
Inclined at an angle
Greater than that
of end generator.

Ξ±

Ξ²
Ξ²= Ξ±

Section Plane Parallel
to end generator.

Parabola
e=1

Hyperbola
e>1
What is eccentricity ?
Conic
section

A

Directrix

P
N

Axis

C

D

V
F

Focus

Vertex
B
eccentricity =

Distance from focus
Distance from directrix

=

PF
PN

=

VF
VC
COMMON DEFINATION OF ELLIPSE, PARABOLA & HYPERBOLA:
These are the loci of points moving in a plane such that the ratio of it’s distances
from a fixed point And a fixed line always remains constant.
The Ratio is called ECCENTRICITY. (E)
A) For Ellipse
E<1
B) For Parabola E=1
C) For Hyperbola E>1
P

C

A

F2

F1

D

B

AB: Major Axis
CD: Minor Axis

PF1+PF2=Constant=AB= Major Axis

SECOND DEFINATION OF AN ELLIPSE:It is a locus of a point moving in a plane
such that the SUM of it’s distances from TWO fixed points
always remains constant.
{And this sum equals to the length of major axis.}
These TWO fixed points are FOCUS 1 & FOCUS 2
Problem: Draw an ellipse by general method, given distance of focus from directrix 50 mm and
eccentricity 2/3. Also draw normal and tangent on the curve at a point 50 mm from the focus.
A

’
10

11

’

9’

8’
7’
6’
5’
4’
3’
2’
1’

1. Draw a vertical line AB of any length as
directrix and mark a point C on it.
2. Draw a horizontal line CD of any length from
point C as axis
3. Mark a point F on line CD at 50 mm from C
4. Divide CF in 5 equal divisions
5. Mark V on 2nd division from F
6. Draw a perpendicular on V and mark a point E
on it at a distance equal to VF
7. Join CE end extend it
8. Mark points 1,2,3…on CF beyond V at uniform
distance, and draw perpendiculars on each of them
so as to intersect extended CE at 1’,2’,3’...

E

C

V

1

F2

3

4

90ΒΊ

NO
RM
AL

TAN
GEN
T

B

90ΒΊ

5

6

7

8

9

10

11

D
Problem: Draw a parabola by general method, given distance of focus from directrix 50 mm.
Also draw normal and tangent on the curve at a point 50 mm from the focus.
1. Draw a vertical line AB of any length as
directrix and mark a point C on it.
2. Draw a horizontal line CD of any length from
point C as axis
3. Mark a point F on line CD at 50 mm from C

A

3’
1’

4’

5’

6’

7’

8’

9’

2’

5. Mark V on mid point of CF
6. Draw a perpendicular on V and mark a point E
on it at a distance equal to VF
7. Join CE end extend it
8. Mark points 1,2,3…on CF beyond V at uniform
distance, and draw perpendiculars on each of them
so as to intersect extended CE at 1’,2’,3’...

E

C

V

90ΒΊ

1 2

F
3 4 5 6 7 8 9

90ΒΊ
MA
OR
N

L

T
EN
NG
TA

B

D
ELLIPSE

Problem 1:Draw ellipse by concentric circle method.
Take major axis 150 mm and minor axis 100 mm long.
Also draw normal and tangent on the curve at a point
25mm above the major axis

3
2

4
C

1
P

F1

1’
No
rm
a

5’

l

O

B

F2

10’

6’
9’

10

5

4’

25mm

A

3’

2’

Ta
ng
en
t

Steps:
1. Draw both axes as perpendicular bisectors
of each other & name their ends as shown.
2. Taking their intersecting point as a center,
draw two concentric circles considering both
as respective diameters.
3. Divide both circles in 12 equal parts &
name as shown.
4. From all points of outer circle draw vertical
lines downwards and upwards respectively.
5.From all points of inner circle draw
horizontal lines to intersect those vertical
lines.
6. Mark all intersecting points properly as
those are the points on ellipse.
7. Join all these points along with the ends of
both axes in smooth possible curve. It is
required ellipse.

BY CONCENTRIC CIRCLE METHOD

7’
6

8’
D
9

7
8
ELLIPSE

BY RECTANGLE METHOD

Problem 2
Draw ellipse by Rectangle method.Take major axis 100 mm and
minor axis 70 mm long. Also draw a normal and a tangent on the
curve at a point 25 mm above the major axis.

D

4’

Tan
gen
t

3’
2’

25mm

A

l

1’
1

2

3

4O

C

No
rma

Steps:
1 Draw a rectangle taking major
and minor axes as sides.
2. In this rectangle draw both
axes as perpendicular bisectors
of each other..
3. For construction, select upper
left part of rectangle. Divide
vertical small side and horizontal
long side into same number of
equal parts.( here divided in four
parts)
4. Name those as shown..
5. Now join all vertical points
1’,2’,3’,4’, to the upper end of
minor axis. And all horizontal
points i.e.1,2,3,4 to the lower end
of minor axis.
6. Then extend C-1 line upto D1’ and mark that point. Similarly
extend C-2, C-3, C-4 lines up to
D-2’, D-3’, & D-4’ lines.
7. Mark all these points properly
and join all along with ends A
and D in smooth possible curve.
Points in the remaining three
quadrants can be marked using
principal of symmetry. Then join
all the points so obtained. It is
required ellipse.

B
PROBLEM 4.
MAJOR AXIS AB & MINOR AXIS CD ARE
100 AMD 70MM LONG RESPECTIVELY
.DRAW ELLIPSE BY ARCS OF CIRLES
METHOD.

ELLIPSE

BY ARCS OF CIRCLE METHOD
As per the definition Ellipse is locus of point P moving in
a plane such that the SUM of it’s distances from two fixed
points (F1 & F2) remains constant and equals to the length
of major axis AB.(Note A .1+ B .1=A . 2 + B. 2 = AB)

STEPS:
1.Draw both axes as usual.Name the
ends & intersecting point
2.Taking AO distance I.e.half major
axis, from C, mark F1 & F2 On AB .
( focus 1 and 2.)
P2
3.On line F1- O taking any distance,
P1
mark points 1,2,3, & 4
4.Taking F1 center, with distance A-1
draw an arc above AB and taking F2
center, with B-1 distance cut this arc.
A
Name the point p1
F1 1
5.Repeat this step with same centers but
taking now A-2 & B-2 distances for
drawing arcs. Name the point p2
P1
6.Similarly get all other P points.
With same steps positions of P can be
P2
located below AB.
7.Join all points by smooth curve to get
an ellipse/

C

P4

P3

P4

P3
P2
P1

2

3

4

O

B

F2
P1

P3

P4

P3

D

P4

P2
ELLIPSE
TANGENT & NORMAL

Problem 13:

TO DRAW TANGENT & NORMAL
TO THE CURVE FROM A GIVEN POINT ( Q )

p4

p3

C

p2
p1

F1

1

2

3

4

O

AL

A

NOR
M

3.

1.
JOIN POINT Q TO F1 & F2
2.
BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL
A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE.

Q

TAN
GE
NT

D

B
F2
Problem 3:Draw ellipse by Oblong method.
Draw a parallelogram of 100 mm and 70 mm long
sides with included angle of 750.Inscribe Ellipse in it.

ELLIPSE

BY OBLONG METHOD

STEPS ARE SIMILAR TO
THE PREVIOUS CASE
(RECTANGLE METHOD)
ONLY IN PLACE OF RECTANGLE,
HERE IS A PARALLELOGRAM.

D
4

4

3

3
2

2
1

1

A

1

2

3

C

4

3

2

1

B
PARABOLA
RECTANGLE METHOD

PROBLEM 7: A BALL THROWN IN AIR ATTAINS 100 M HIEGHT
AND COVERS HORIZONTAL DISTANCE 150 M ON GROUND.
Draw the path of the ball (projectile)-

STEPS:
1.Draw rectangle of above size and
divide it in two equal vertical parts
2.Consider left part for construction.
Divide height and length in equal
number of parts and name those
1,2,3,4,5& 6
3.Join vertical 1,2,3,4,5 & 6 to the
top center of rectangle
4.Similarly draw upward vertical
lines from horizontal1,2,3,4,5
And wherever these lines intersect
previously drawn inclined lines in
sequence Mark those points and
further join in smooth possible curve.
5.Repeat the construction on right side
rectangle also.Join all in sequence.
This locus is Parabola.
.

Scale 1cm = 10m.

6
5
4
3
2
1

1’

2’

3’

4’

5’

6’
Draw a parabola by tangent(or triangulation) method given base 7.5m and axis 4.5m
Take scale 1cm = 0.5m

O
10
9

1’

8

2’

7
6
5

4.5m

3’
4’

F

4

5’
6’

3

7’

2

4.5m
8’

1

9’
10’

A

E
7.5m

B
Problem 51: A fountain jet discharges water from ground level at an inclination of 45ΒΊ to
the ground. The jet travels a horizontal distance of 7.5m from the point of discharge and
falls on the ground. Trace the path of the jet. Name the curve.
As the jet will be a projectile so its path will be parabolic. The angle of jet with the ground is the angle of tangent on the curve at the point of
discharge. First we will consider a scale to accommodate 7.5 m on the ground. That can be done by considering 1cm= 0.5 m.

5

O

4

1’

3

2’

2

3’

1

A

4’

45ΒΊ

45ΒΊ

7.5m

5’

B
HYPERBOLA

Problem No.10: Point P is 40 mm and 30 mm from horizontal
and vertical axes respectively.Draw Hyperbola through it.

THROUGH A POINT
OF KNOWN CO-ORDINATES

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
after P-1, 2,3,4 etc.
4) Join 1-2-3-4 points
to pole O. Let them cut
part [P-B] also at 1,2,3,4
points.
5) From horizontal
1,2,3,4 draw vertical
lines downwards and
6) From vertical 1,2,3,4
points [from P-B] draw
horizontal lines.
40 mm
7) Line from 1
horizontal and line from
1 vertical will meet at
P1.Similarly mark P2, P3,
P4 points.
8) Repeat the procedure
by marking four points
on upward vertical line
from P and joining all
those to pole O. Name
this points P6, P7, P8 etc.
and join them by smooth

1’

2’

1

2

P
3’
4’
5’

O
30 mm

3

4

5
Arc of circle Method
Problem 14: Two points A and B are 50 mm apart. A point P moves in a plane in such a
way that the difference of its distance from A and B is always constant and equal to 20
mm. Draw the locus of point P.
Draw a line and mark two points A & B on it
at a distance of 50 mm.
Mark O as mid point of AB.
Mark two points V1 and V2 at 10 mm on
either side of O.
Mark points 1, 2,3 on the right of Bat any
distances.
o
A

V1

V2

10 10
50

B

1

2

3

As per the definition Hyperbola is locus of point P
moving in a plane such that the difference of it’s
distances from two fixed points (F1 & F2) remains
constant and equals to the length
of transverse axis V1 V2.
Take V11 as radius and A as centre and
draw an arc on the right side of A.
Take V21 as radius and B as centre and
draw an arc on the left side of B so as to
intersect the previous arc.
Repeat the step with V12, V22 as radii and
V13, V23 as radii respectively.
Repeat the same steps on the other side to
draw the second hyperbola.
ELLIPSE
TANGENT & NORMAL

Problem 14:

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.

ELLIPSE

A

DIRECTRIX

TO DRAW TANGENT & NORMAL
TO THE CURVE
FROM A GIVEN POINT ( Q )

T

(vertex) V

F ( focus)
900

N

N

Q

B

T
PARABOLA
TANGENT & NORMAL

Problem 15:
TO DRAW TANGENT & NORMAL
TO THE CURVE
FROM A GIVEN POINT ( Q )

T

PARABOLA

A

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.

VERTEX V

900

F
( focus)
N

Q

B

N

T
HYPERBOLA
TANGENT & NORMAL

Problem 16
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 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.

A

T
(vertex)

F ( focus)

V

900
N

N

Q

B
T

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Curve1

  • 1. ENGINEERING CURVES Part- I {Conic Sections} ELLIPSE PARABOLA HYPERBOLA 1.Concentric Circle Method 1.Rectangle Method 2.Rectangle Method 2 Method of Tangents ( Triangle Method) 1.Rectangular Hyperbola (coordinates given) 3.Oblong Method 4.Arcs of Circle Method 3.Basic Locus Method (Directrix – focus) 2 Rectangular Hyperbola (P-V diagram - Equation given) 3.Basic Locus Method (Directrix – focus) 5.Rhombus Metho 6.Basic Locus Method (Directrix – focus) Methods of Drawing Tangents & Normals To These Curves.
  • 3. CONIC SECTIONS ELLIPSE, PARABOLA AND HYPERBOLA ARE CALLED CONIC SECTIONS BECAUSE THESE CURVES APPEAR ON THE SURFACE OF A CONE WHEN IT IS CUT BY SOME TYPICAL CUTTING PLANES. OBSERVE ILLUSTRATIONS GIVEN BELOW.. Ellipse Ξ² e<1 Ξ± Ξ± Ξ² Ξ²< Ξ± Ξ²> Ξ± Section Plane Through all the Generators Section Plane Inclined at an angle Greater than that of end generator. Ξ± Ξ² Ξ²= Ξ± Section Plane Parallel to end generator. Parabola e=1 Hyperbola e>1
  • 4. What is eccentricity ? Conic section A Directrix P N Axis C D V F Focus Vertex B eccentricity = Distance from focus Distance from directrix = PF PN = VF VC
  • 5. COMMON DEFINATION OF ELLIPSE, PARABOLA & HYPERBOLA: These are the loci of points moving in a plane such that the ratio of it’s distances from a fixed point And a fixed line always remains constant. The Ratio is called ECCENTRICITY. (E) A) For Ellipse E<1 B) For Parabola E=1 C) For Hyperbola E>1
  • 6. P C A F2 F1 D B AB: Major Axis CD: Minor Axis PF1+PF2=Constant=AB= Major Axis SECOND DEFINATION OF AN ELLIPSE:It is a locus of a point moving in a plane such that the SUM of it’s distances from TWO fixed points always remains constant. {And this sum equals to the length of major axis.} These TWO fixed points are FOCUS 1 & FOCUS 2
  • 7. Problem: Draw an ellipse by general method, given distance of focus from directrix 50 mm and eccentricity 2/3. Also draw normal and tangent on the curve at a point 50 mm from the focus. A ’ 10 11 ’ 9’ 8’ 7’ 6’ 5’ 4’ 3’ 2’ 1’ 1. Draw a vertical line AB of any length as directrix and mark a point C on it. 2. Draw a horizontal line CD of any length from point C as axis 3. Mark a point F on line CD at 50 mm from C 4. Divide CF in 5 equal divisions 5. Mark V on 2nd division from F 6. Draw a perpendicular on V and mark a point E on it at a distance equal to VF 7. Join CE end extend it 8. Mark points 1,2,3…on CF beyond V at uniform distance, and draw perpendiculars on each of them so as to intersect extended CE at 1’,2’,3’... E C V 1 F2 3 4 90ΒΊ NO RM AL TAN GEN T B 90ΒΊ 5 6 7 8 9 10 11 D
  • 8. Problem: Draw a parabola by general method, given distance of focus from directrix 50 mm. Also draw normal and tangent on the curve at a point 50 mm from the focus. 1. Draw a vertical line AB of any length as directrix and mark a point C on it. 2. Draw a horizontal line CD of any length from point C as axis 3. Mark a point F on line CD at 50 mm from C A 3’ 1’ 4’ 5’ 6’ 7’ 8’ 9’ 2’ 5. Mark V on mid point of CF 6. Draw a perpendicular on V and mark a point E on it at a distance equal to VF 7. Join CE end extend it 8. Mark points 1,2,3…on CF beyond V at uniform distance, and draw perpendiculars on each of them so as to intersect extended CE at 1’,2’,3’... E C V 90ΒΊ 1 2 F 3 4 5 6 7 8 9 90ΒΊ MA OR N L T EN NG TA B D
  • 9. ELLIPSE Problem 1:Draw ellipse by concentric circle method. Take major axis 150 mm and minor axis 100 mm long. Also draw normal and tangent on the curve at a point 25mm above the major axis 3 2 4 C 1 P F1 1’ No rm a 5’ l O B F2 10’ 6’ 9’ 10 5 4’ 25mm A 3’ 2’ Ta ng en t Steps: 1. Draw both axes as perpendicular bisectors of each other & name their ends as shown. 2. Taking their intersecting point as a center, draw two concentric circles considering both as respective diameters. 3. Divide both circles in 12 equal parts & name as shown. 4. From all points of outer circle draw vertical lines downwards and upwards respectively. 5.From all points of inner circle draw horizontal lines to intersect those vertical lines. 6. Mark all intersecting points properly as those are the points on ellipse. 7. Join all these points along with the ends of both axes in smooth possible curve. It is required ellipse. BY CONCENTRIC CIRCLE METHOD 7’ 6 8’ D 9 7 8
  • 10. ELLIPSE BY RECTANGLE METHOD Problem 2 Draw ellipse by Rectangle method.Take major axis 100 mm and minor axis 70 mm long. Also draw a normal and a tangent on the curve at a point 25 mm above the major axis. D 4’ Tan gen t 3’ 2’ 25mm A l 1’ 1 2 3 4O C No rma Steps: 1 Draw a rectangle taking major and minor axes as sides. 2. In this rectangle draw both axes as perpendicular bisectors of each other.. 3. For construction, select upper left part of rectangle. Divide vertical small side and horizontal long side into same number of equal parts.( here divided in four parts) 4. Name those as shown.. 5. Now join all vertical points 1’,2’,3’,4’, to the upper end of minor axis. And all horizontal points i.e.1,2,3,4 to the lower end of minor axis. 6. Then extend C-1 line upto D1’ and mark that point. Similarly extend C-2, C-3, C-4 lines up to D-2’, D-3’, & D-4’ lines. 7. Mark all these points properly and join all along with ends A and D in smooth possible curve. Points in the remaining three quadrants can be marked using principal of symmetry. Then join all the points so obtained. It is required ellipse. B
  • 11. PROBLEM 4. MAJOR AXIS AB & MINOR AXIS CD ARE 100 AMD 70MM LONG RESPECTIVELY .DRAW ELLIPSE BY ARCS OF CIRLES METHOD. ELLIPSE BY ARCS OF CIRCLE METHOD As per the definition Ellipse is locus of point P moving in a plane such that the SUM of it’s distances from two fixed points (F1 & F2) remains constant and equals to the length of major axis AB.(Note A .1+ B .1=A . 2 + B. 2 = AB) STEPS: 1.Draw both axes as usual.Name the ends & intersecting point 2.Taking AO distance I.e.half major axis, from C, mark F1 & F2 On AB . ( focus 1 and 2.) P2 3.On line F1- O taking any distance, P1 mark points 1,2,3, & 4 4.Taking F1 center, with distance A-1 draw an arc above AB and taking F2 center, with B-1 distance cut this arc. A Name the point p1 F1 1 5.Repeat this step with same centers but taking now A-2 & B-2 distances for drawing arcs. Name the point p2 P1 6.Similarly get all other P points. With same steps positions of P can be P2 located below AB. 7.Join all points by smooth curve to get an ellipse/ C P4 P3 P4 P3 P2 P1 2 3 4 O B F2 P1 P3 P4 P3 D P4 P2
  • 12. ELLIPSE TANGENT & NORMAL Problem 13: TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) p4 p3 C p2 p1 F1 1 2 3 4 O AL A NOR M 3. 1. JOIN POINT Q TO F1 & F2 2. BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE. Q TAN GE NT D B F2
  • 13. Problem 3:Draw ellipse by Oblong method. Draw a parallelogram of 100 mm and 70 mm long sides with included angle of 750.Inscribe Ellipse in it. ELLIPSE BY OBLONG METHOD STEPS ARE SIMILAR TO THE PREVIOUS CASE (RECTANGLE METHOD) ONLY IN PLACE OF RECTANGLE, HERE IS A PARALLELOGRAM. D 4 4 3 3 2 2 1 1 A 1 2 3 C 4 3 2 1 B
  • 14. PARABOLA RECTANGLE METHOD PROBLEM 7: A BALL THROWN IN AIR ATTAINS 100 M HIEGHT AND COVERS HORIZONTAL DISTANCE 150 M ON GROUND. Draw the path of the ball (projectile)- STEPS: 1.Draw rectangle of above size and divide it in two equal vertical parts 2.Consider left part for construction. Divide height and length in equal number of parts and name those 1,2,3,4,5& 6 3.Join vertical 1,2,3,4,5 & 6 to the top center of rectangle 4.Similarly draw upward vertical lines from horizontal1,2,3,4,5 And wherever these lines intersect previously drawn inclined lines in sequence Mark those points and further join in smooth possible curve. 5.Repeat the construction on right side rectangle also.Join all in sequence. This locus is Parabola. . Scale 1cm = 10m. 6 5 4 3 2 1 1’ 2’ 3’ 4’ 5’ 6’
  • 15. Draw a parabola by tangent(or triangulation) method given base 7.5m and axis 4.5m Take scale 1cm = 0.5m O 10 9 1’ 8 2’ 7 6 5 4.5m 3’ 4’ F 4 5’ 6’ 3 7’ 2 4.5m 8’ 1 9’ 10’ A E 7.5m B
  • 16. Problem 51: A fountain jet discharges water from ground level at an inclination of 45ΒΊ to the ground. The jet travels a horizontal distance of 7.5m from the point of discharge and falls on the ground. Trace the path of the jet. Name the curve. As the jet will be a projectile so its path will be parabolic. The angle of jet with the ground is the angle of tangent on the curve at the point of discharge. First we will consider a scale to accommodate 7.5 m on the ground. That can be done by considering 1cm= 0.5 m. 5 O 4 1’ 3 2’ 2 3’ 1 A 4’ 45ΒΊ 45ΒΊ 7.5m 5’ B
  • 17. HYPERBOLA Problem No.10: Point P is 40 mm and 30 mm from horizontal and vertical axes respectively.Draw Hyperbola through it. THROUGH A POINT OF KNOWN CO-ORDINATES 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 after P-1, 2,3,4 etc. 4) Join 1-2-3-4 points to pole O. Let them cut part [P-B] also at 1,2,3,4 points. 5) From horizontal 1,2,3,4 draw vertical lines downwards and 6) From vertical 1,2,3,4 points [from P-B] draw horizontal lines. 40 mm 7) Line from 1 horizontal and line from 1 vertical will meet at P1.Similarly mark P2, P3, P4 points. 8) Repeat the procedure by marking four points on upward vertical line from P and joining all those to pole O. Name this points P6, P7, P8 etc. and join them by smooth 1’ 2’ 1 2 P 3’ 4’ 5’ O 30 mm 3 4 5
  • 18. Arc of circle Method Problem 14: Two points A and B are 50 mm apart. A point P moves in a plane in such a way that the difference of its distance from A and B is always constant and equal to 20 mm. Draw the locus of point P. Draw a line and mark two points A & B on it at a distance of 50 mm. Mark O as mid point of AB. Mark two points V1 and V2 at 10 mm on either side of O. Mark points 1, 2,3 on the right of Bat any distances. o A V1 V2 10 10 50 B 1 2 3 As per the definition Hyperbola is locus of point P moving in a plane such that the difference of it’s distances from two fixed points (F1 & F2) remains constant and equals to the length of transverse axis V1 V2. Take V11 as radius and A as centre and draw an arc on the right side of A. Take V21 as radius and B as centre and draw an arc on the left side of B so as to intersect the previous arc. Repeat the step with V12, V22 as radii and V13, V23 as radii respectively. Repeat the same steps on the other side to draw the second hyperbola.
  • 19. ELLIPSE TANGENT & NORMAL Problem 14: 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. ELLIPSE A DIRECTRIX TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) T (vertex) V F ( focus) 900 N N Q B T
  • 20. PARABOLA TANGENT & NORMAL Problem 15: TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) T PARABOLA A 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. VERTEX V 900 F ( focus) N Q B N T
  • 21. HYPERBOLA TANGENT & NORMAL Problem 16 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 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. A T (vertex) F ( focus) V 900 N N Q B T