# Hyperbola Directrix e>1 O: center F1, F2: foci V1, V2: vertices

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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

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’

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

Hyperbola-rectangle method
1’ 2’ 2 1 Base Height of hyperbola Axis height

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 )

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

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

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

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

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’

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’

To obtain the edge view of a plane
c1 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’

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)

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