How Conic Section is Formed

Conic Sections

Cone is formed when a right angled triangle with an apex and angle e is rotated about its altitude as the axis. The length or height of the cone is equal to the altitude of the triangle and the radius of the base of the cone is equal to the base of the triangle. The apex angle of the cone is 2 e (Fig. 1a).

When a cone is cut by a plane, the curve formed along the section is known as a conic. For this purpose, the cone may be cut by different section planes (Fig. 1b) and the conic sections obtained are shown in Fig. 1c, d, and e.

Fig. 1 (a) & (b)
Fig. 1 (c), (d) & (e)

Circle

When a cone is cut by a section plane A-A making an angle a = 90° with the axis, the section obtained is a circle. (Fig 1a)

Ellipse

When a cone is cut by a section plane B-B at an angle, a more than half of the apex angle i.e., θ and less than 90°, the curve of the section is an ellipse. Its size depends on the angle α and the distance of the section plane from the apex of the cone.

Parabola

If the angle a is equal to θ i.e., when the section plane C-C is parallel to the slant side of the cone. the curve at the section is a parobola. This is not a closed figure like circle or ellipse. The size of the parabola depends upon the distance of the section plane from the slant side of the cone.

Hyperbola

If the angle a is less than θ (section plane D-D), the curve at the section is hyperbola. The curve of intersection is hyperbola, even if α = θ, provided the section plane is not passing through the apex of the cone. However if the section plane passes through the apex, the section produced is an isosceles triangle.

Conic Sections as Loci of a Moving Point

A conic section may be defined as the locus of a point moving in a plane such that the ratio of its distance from a fixed point (Focus) and fixed straight line (Directrix) is always a constant. The ratio is called eccentricity. The line passing through the focus and perpendicular to the directrix is the axis of the curve. The point at which the conic section intersects the axis is called the vertex or apex of the curve.

The eccentricity value is less than 1 for ellipse, equal to I for parabola and greater than 1 for hyperbola (F ig. 2).

Fig. 2

To draw a parabola with the distance of the focus from the directrix at 50mm (Eccentricity method Fig. 3).

1. Draw the axis AB and the directrix CD at right angles to it:
2. Mark the focus F on the axis at 50mm.
3. Locate the vertex V on AB such that AV = VF
4. Draw a line VE perpendicular to AB such that VE = VF\
5. Join A,E and extend. Now, VE/VA = VF/VA = 1, the eccentricity.
6. Locate number of points 1,2,3, etc., to the right of V on the axis, which need not be equidistant.
7. Through the points 1,2,3, etc., draw lines perpendicular to the axis and to meet the line AE extended at 1′,2′,3′ etc.
8. With centre F and radius 1-1, draw arcs intersecting the line through I at P I and P11.
9. Similarly, lolcate the points P2, P21, P3, P31, etc., on either side of the axis. Join the points by smooth curve, forming the required parabola.

Construction of a Parabola -Eccentricity Method
Fig. 3 Construction of a Parabola -Eccentricity Method

To draw a normal and tangent through a point 40mm from the directrix.

To draw a tangent and normal to the parabola. locate the point M which is at 40 mm from the directQx. Thenjoin M to F and draw a line through F, perpendicular to MF to meet the directrix at T. The line joining T and M and extended is the tangent and a line NN, through M and perpendicular to TM is the normal to the curve.

To draw an Ellipse with eccentricity equal to 2/3 for the above problem (Fig. 4).

Construction is similar to the one in Fig. 3 to draw an ellipse including the tangent and normal. only the eccentricity is taken as 2/3 instead of one.

Fig. 4

Draw a hyperbola with eccentricity equal to 3/2 for.the above problem (Fig. 5).

The construction of hyperobola is similar to the above problems except that the eccentricity ratio VF/VA = 3/2 in this case.

Note: The ellipse ,is a closed curve and has two foci and two directrices. A hyperbola is an open curve.

Fig. 5

Other Methods of Construction of Ellipse

Given the dimensions of major and minor axes the ellipse can be drawn by, (i) Foci method, (ii) Oblong method, (iii) Concentric circle method and (iv) Trammel method.

To draw an ellipse with major and minor axes equal to 120 mm and 80 mm respectively. Definition of Ellipse (Fig. 6)

Ellipse is a curve traced by a point moving such that the sum of its distances from the two fixed points, foci, is constant and equal to the major axis.

Properties of an Ellipse
Fig. 6 Properties of an Ellipse

Referring Fig. 6, F1 , and F2 are the two foci, AB is the major axis and CD is the minor axis. As per the difmition, PF1 + PF2 = CF1 + CF2 = QF1 + QF2 = AB. It may also be noted that CF1 = CF2 = 1/2 AB (Major axis)

Construction

1. Foci Method (Fig. 7)

Construction of Ellipse - Foci Method
Fig. 7 Construction of Ellipse – Foci Method

1. Draw the major (AB) and ninor (CD) axes and locate the centre O.
2. Locate the foci F1 and F2 by taking a radius equal to 60 mm (1/2 of AB) and cutting AB at F1 P1 and F2 with C as the centre.
3. Mark a number of points 1,2,3 etc., between F1 and O, which need not be equi-distance.
4. With centres F1 and F2 and radii A1 and B1 respectively, draw arcs intersecting at the points P1 and P1 .
5. Again with centres F1 and F2 and radii Bland A1 respectively, draw arcs intersecting at the points Q1 and Q1 .
6. Repeat the steps 4 and 5 with the remaining points 2,3,4 etc., and obtain additional points on the curve.
Join the points by a smooth curve, forming the required ellipse.

To mark a Tangent and Normal to the ellipse at any point, say M on it, join the foci F1 and F2 with M and extend F2M to E and bisect the angle <EMF1‘ The bisector TT represents the required tangent and a line NN drawn through M and perpendicular to TT is the normal to the ellipse.

2. Oblong Method (Fig. 8)

Oblong Method
Fig. 8 Oblong Method

1. Draw the major and minor axes AB and CD and locate the centre O.
2. Draw the rectangle KLMN passing through A,D,B,C.
3. Divide AO and AN into same number of equal parts, say 4.
4. Join C with the points 1′,2′,3′ .
5. JoinD with the points 1,2,3 and extend till they meet the lines C1, C2, C3 “respectively at P1, P2 and P3.
6. Repeat steps 3 to 5 to obtain the points in the remaining three quadrants.
7. Join the points by a smooth curve forming the required ellipse.

To draw an ellipse passing through any three given points not in a line.
Construction (Fig. 9)

Fig. 9

1. Locate the given points A,B and C
2. Join A and B (which is longer than AC and BC) and locate its centre. This becomes the
major axis of the ellipse.
3. Draw CO and extend it to D such that CO = OD and CD is the minor axis of the ellipse.
4. Draw the parallelogram KLMN, Passing through A,D,B and C.
5. Follow the”steps given is Fig 8 and obtain the points on the curve.
6. Join the points by a smooth curve, forming the required ellipse.

3. Concentric Circles Method (Fig. 10)

1. Draw the major and minor axes AB and CD and locate the centre O.
2. With centre 0 and major axis and minor axes as diameters, draw two concentric circles.
3. Divide both the circles into equal number of parts, say 12 and draw the radial lines.
4. Considering the radial line 0-1‘ -1, draw a horizontal line from I’ to meet the vertical line from 1 at P1
5. Repeat the steps 4 and obtain other points P2, P3, etc.
6. Join the points by a smooth curve forming the required ellipse.

Concetric Circle Method
Fig. 10 Concetric Circle Method

4. Trammel Method (Fig. 11)

  1. Draw the major and minor axes AB and CD and then locate the centre O.
  2. Take.a strip of stiff paper and prepare a trammel as shown. Mark the semi-major and semi-minor axes PR and PQ on the trammel.
  3. Position the trammel so that the points R and Q lie respectively on the minor and major axes. As a rule, the third point P will always lie on the ellipse required.
  4. Keeping R on the minor axis and Q on the major axis, move the trammel to Qther position and locate other points on the curve.
  5. Join the points by a smooth curve forming the required ellipse.
Trammel Method
Fig. 11 Trammel Method

Other Methods of Constructing Parabola

To draw a parabola with 70 mm as base and 30 mm as the length of the axis.

1. Tangent Method (Fig. 12)

Tangent Method
Fig. 12 Tangent Method

 

1. Draw the base AB and locate its mid-point C.
2. Through C, draw CD perpendicular to AB forning the axis
3. Produce CD to E such that DE = CD
4. Join E-A and E-B. These are the tangents to the parabola at A and B.
5. Divide AE and BE into the same number of equal parts and number the points as shown.
6. Join 1-1′ ,2- 2′ ,3- 3′ , etc., forming the tangents to the required parabola.
7. A smooth curve passing through A, D and B and tangential to the above lines is the required parabola.

Note: To draw a tangent to the curve at a point, say M on it, draw a horizontal through M, meeting the axis at F. mark G on the extension of the axis such that DG = FD. Join G, M and extend, forming the tangent to the curve at M.

2. Rectangle Method (Fig. 13)

Construction of Parabola
Fig. 13 Construction of Parabola

1. Draw the base AB and axis CD such that CD is perpendicular bisector to AB.
2. Construct a rectangle ABEF, passing through C.
3. Divide AC and AF into the same number of equal parts and number the points ‘as shown.
4. Join 1,2 and 3 to D.
5. Through 1′,2′ and 3’, draw lines parallel to the axis, intersecting the lines ID, 2D and 3D at P1‘ P2 and P3 respectively.
6. Obtain the points P1, P2 and P3, which are symmetrically placed to P1‘ P2 and P3 with respect to the axis CD.
7. Join the points by a smooth curve forming the required parabola.

Note: Draw a tangent at M following the method indicated in Fig. 12.

Method of constructing a hyperbola, given the foci and the distance between the vertices. (Fig 14)

A hyperbola is a curve generated by a point moving such that the difference of its distances from two fixed points called foci is always constant and equal to the distance between the vertices of the two branches of hyperbola. This distance is also known as the major a.xis of the hyperbola.

Properties of Hyperbola
Fig. 14 Properties of Hyperbola

Referipg Fig. 14, the difference between P1F1∼P1F2= P2F2∼P2F1 = V1V2 (major axis)

The axesAB and CD are known as transverse and conjugate axes of the hyperbola. The curve has two branches which are symmetric about the conjugate axis.

Problem : Construct a hyperbola with its foci 70 mm apart and the major axis (distance between the vertices)as 40 mm. Draw a tangent to the curve at a point 20 mm from the focus.

Construction (Fig. 15)

1. Draw the transverse and conjugate axes AB and CD of the hyperbola and locate F1 and F2‘ the foci and V1 and V2‘ the vertices.
2. Mark number of points 1,2,3 etc., on the transverse axis, which need not be equi-distant.

Construction of a Hyperbola
Fig. 15 Construction of a Hyperbola

3. With centre F1 and radius V11, draw arcs on either side of the transverse a.xis.
4. With centre F2 and radius V21, draw arcs intersecting the above arcs at P1, and P1,
5. With centre F2 and radius V11, draw arcs on either side of the transverse axis.
6. With centre F1 and radius V2 1, draw arcs intersecting the above arcs at Q1‘ Q11
7. Repeat the steps 3 to 6 and obtain other points P2‘ P12‘ etc. and Q2, Q12‘ etc.
8. Join the points P1,P2, P3, P’1,P’2,P’3 and Q1,Q2,Q3, Q’1,Q’2,Q’3 forming the two branches of hyperbola.

Note: To draw a tangent to the hyperbola, locate the point M which is at 20mm from the focus say F2‘ Then, join M to the foci F1 and F2‘ Draw a line IT, bisecting the <F1 MF2 forming the required tangent at M.

To draw the asymptotes to the given hyperbola

Lines passing through the centre and tangential to the curve at infinity are known as asymptotes.

Construction (16)

1. Through the vertices V1 and V2 draw perpendiculars to the transverse axis.
2. With centre O and radius OF1 = (OF2)’ draw a circle meeting the above lines at P, Q and R,S.
3. Join the points P,O,R and S,O,Q and extend, forming the asymptotes to the hyperbola.

Note: The circle drawn with O as centre and V1 V2 as diameters is known as auxiliary circle. Asymptotes intersect the auxiliary circle on the directrix. Thus. D1‘ D1 and D2D2 are the two directrices for the two branches of hyperbola.

Drawing asymptotes to a hyperbola
Fig. 16 Drawing asymptotes to a hyperbola

Rectangular Hyperbola

When the asymptotes to the hyperbola intersect each other at right angles, the curve is known as a rectangular hyperbola.

Application of Conic Curves

An ellipsoid is generated by rotating an ellipse about its major axis. An ellipsoidal surface is used as a head-lamp reflector. The light source (bulb) is placed at the first focus F1 (Fig. 17). This works effectively, if the second focus F2 is at a sufficient distance from the first focus. Thus, the light rays reflecting from the surface are almost parallel to each other.

Ellipsoidal Reflector
Fig. 17 Ellipsoidal Reflector

Parabolic Curve

The parabolic curve finds its application for reflecting surfaces of light, Arch forms, cable forms in suspension bridges, wall brickets of uniform strength, etc.

The paraboloid reflector may be used as a solar heater. When it is properly adjusted, the sun rays emanating from infinite distance, concentrate at the focus and thus produce more heat. The wall bracket of parabolic shape exhibits equal bending strength at all sections (Fig. 18)

Wall bracket of uniform strength
Fig. 18 Wall bracket of uniform strength

Hyperbola

A rectangular hyperboler is a graphical representation of Boyes law, PV=Constant. This curve also finds its application in the design of water channels.

Problem : Draw an ellipse with mojor axis 120 mm and minor axis 80 mm. Determine the eccentricity and the distance between the directrices.

Construction (Fig. 19)

Fig. 19

Eccentricity e = V1 F1 / V1 A = V1 F2 / V1B
therefore V1 F2-V1F1 / V1B-V1A = F1 F2/ V1V2
From the triangle F1 CO
OC = 40 mm (half of minor axis)
F1F2 = 60 mm (half of major axis)

    \[Thus F_1 0 =\sqrt{60^2 - 40^2} = 44.7mm\]

Hence F1F2 = 2F1O = 89.4 mm

    \[\text{on substitution e} = \frac{89.4}{120} = 0.745 \]

Also,eccentricity e = V1 V2 /AB, Hence, the distance between the directrices AB = V1V2 /e = 161mm

Problem : A fountain jet is dicharged from the ground level at an inclination of 45°. The jet travels a horizontal distance of 10m from the point of discharge and falls on the ground. ‘Trace the path of the jet.

Construction (Fig. 20)

Fig. 20

1. Draw the base AB of 10m long and locate its mid-point C.
2. Through C draw a line perpendicular to AB forming the axis.
3. Through A and B, draw lines at 45°, to the base intersecting the axis at D.
4. Divide AD and BD irtto the same number of equal parts and number the points as shown.
5. Join 1-1′ , 2- 2′ ,3- 3′ etc., forming the tangents to the required path of jet.
6. A smooth curve passing through A and B and tangential to the above lines is the required path of the jet which is parabolic in shape.

Problem : A stone is thrown from a building of 7 m high and at its highest flight it just crosses a plam tree 14 m high. Trace the path of the stone, if the distance between the building and the tree measured along the ground is 3.5 m.

Construction (Fig. 21)

Fig. 21

1. Draw lines AB and OT, representing the building and plam tree respectively, 3.5 m apart and above the ground level.
2. Locate C and D on the horizontal line through B such that CD=BC=3.5 and complete the rectangle BDEF.
3. Inscribe the parabola in the rectangle BDEF, by rectangular method.
4. Draw the path of the stone till it reaches the ground (H) extending the principle of rectangle method.

Support Us By Sharing

Author: Aliva Tripathy

Taking out time from a housewife life and contributing to AxiBook is a passion for me. I love doing this and gets mind filled with huge satisfaction with thoughtful feedbacks from you all. Do love caring for others and love sharing knowledge more than this.

Leave a Reply

Your email address will not be published. Required fields are marked *

*

code