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Hyperbola 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.

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Presentation on theme: "Hyperbola 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."— Presentation transcript:

1 Hyperbola 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. F2F1 Directrix a ae Asymptote O V2V1 b When the asymptotes are perpendicular it is a called a rectangular hyperbola. Hyperbola Axis

2 2 HYPERBOLA DIRECTRIX-FOCUS METHOD Draw an ellipse, focus is 50 mm from the directrix and the eccentricity is 3/2 F1 ( focus) DIRECTRIX 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=2/3 THEREFORE P1 AND P1’ LIE ON THE HYPERBOLA

3 P O 40 mm 30 mm 1’ 2’ 3’ 4’5’ HYPERBOLA 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 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’ P 1.Similarly mark P 2, P 3, P 4 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

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

5 VOLUME:( M 3 ) PRESSURE ( Kg/cm 2 ) HYPERBOLA P-V DIAGRAM 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. Form a table giving few more values of P & V P V = C ============ 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.

6 D F1F1 F2F A B C p1p1 p2p2 p3p3 p4p4 O Q TANGENT NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE AT A GIVEN POINT ( Q ) 1.JOIN POINT Q TO F 1 & F 2 2.BISECT ANGLE F 1 Q F 2 THE ANGLE BISECTOR IS NORMAL 3.A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE. ELLIPSE TANGENT & NORMAL Problem :

7 ELLIPSE TANGENT & NORMAL F ( focus) DIRECTRIX V ELLIPSE (vertex) A B T T N N Q 90 0 TO DRAW TANGENT & NORMAL TO THE CURVE AT A GIVEN POINT ( Q ) 1.JOIN POINT Q TO F. 2.CONSTRUCT 90 0 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. Problem :

8 A B PARABOLA VERTEX F ( focus) V Q T N N T 90 0 TO DRAW TANGENT & NORMAL TO THE CURVE AT A GIVEN POINT ( Q ) 1.JOIN POINT Q TO F. 2.CONSTRUCT 90 0 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. PARABOLA TANGENT & NORMAL Problem :

9 F ( focus) V (vertex) A B HYPERBOLA TANGENT & NORMAL QN N T T 90 0 TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) 1.JOIN POINT Q TO F. 2.CONSTRUCT 90 0 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. Problem 16

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

11 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

12 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: a 1, b 1, c 1 from x 1 y 1 = a’, b’, c’ from xy respectively Edge view of the plane xy c1c1 a1a1 b1b1 x1x1 y1y1

13 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 y c2c2 a2a2 b2b2 c1c1 a1a1 b1b1 x1x1 y1y1  True shape and dimensions of the plane True shape is the auxiliary view obtained from the edge view Edge view of planec3c3 a3a3 b3b3 c4c4 a4a4 b4b4 Edge view of the plane  is the angle of the plane with the HP


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