# Mathematics.

## Presentation on theme: "Mathematics."— Presentation transcript:

Mathematics

PARABOLA - SESSION 1

Session Objectives Definition of Conic Section Eccentricity
Definition of Special Points Standard Form of parabola General Form of parabola Algorithm for finding special points/ lines Condition for Second degree equation to represent different conic sections

Definition of Conic section
Geometrical Definition Cross section formed when right circular cone is intersected by a plane Axis Generator

Circle Circle If plane is perpendicular to the axis
Geometrical Definition Cross section formed when right circular cone is intersected by a plane Circle If plane is perpendicular to the axis

Ellipse Ellipse If plane is not perpendicular to the axis
Geometrical Definition Cross section formed when right circular cone is intersected by a plane Ellipse If plane is not perpendicular to the axis Does not pass through base

Parabola Parabola If plane is parallel to the generator
Geometrical Definition Cross section formed when right circular cone is intersected by a plane Parabola If plane is parallel to the generator

Hyperbola Hyperbola Two similar cones Plane parallel to the axis
Geometrical Definition Cross section formed when right circular cone is intersected by a plane Hyperbola Two similar cones Plane parallel to the axis

Class Exercise Point Pair of straight lines
Are the following be a conic section? Point Pair of straight lines If yes, how they can be generated by intersection of cone(s) and plane.

Class Exercise

Locus Definition Locus Definition Locus of a point moves such that
Ratio of its distance from a fixed point & from a fixed line is constant Fixed Line P N S Fixed Point Ratio - Eccentricity Fixed Point - Focus Fixed Line - Line of Directrix

Eccentricity and Shapes of Conic Section
e = 1 : Parabola e < 1 : Ellipse e = 0 : Circle e > 1 : Hyperbola

Special Points / Lines Axis :
Line through Focus and perpendicular to line of directrix Vertex : Meeting point of Curve and axis Directrix N P S Focus Vertex Axis

Special Points / Lines Double Ordinate :
Line segment joint two points on a conic for one particular value of abscissa Latus rectum : Double ordinate passing through Focus Directrix N P S Focus Axis Vertex

Standard Form of Parabola
e =1 Axis is x- axis , y = 0 Vertex - ( 0,0) Focus - ( a,0) Directrix N P S Focus Axis Vertex V As e = 1 , SV = VV1 Let P be (  , ) V1

Standard Form of Parabola
e =1 Directrix N P S Focus Axis Vertex V V1

Standard Form of Parabola- Special Point / lines
Focus : ( a,0) , Vertex : ( 0,0) Axis : y = 0 , Directrix : x = – a Length of Latus rectum : Eq. Of SLL’ : x = a Directrix N P S Focus Axis Vertex V V1 P.O.I of this line and Parabola : y2 = 4a (a) L L’

Standard Form of Parabola- Special Point / lines
Focus : ( -a,0) , Vertex : ( 0,0) Axis : y = 0 , Directrix : x =–(– a) Length of Latus rectum : Eq. Of SLL’ : x = –a N S Focus Vertex V Directrix P Axis V1 L L’ P.O.I of this line and Parabola : y2 = – 4a (–a)

Standard Form of Parabola- Special Point / lines
Focus : ( 0,a) , Vertex : ( 0,0) Axis : x = 0 , Directrix : y =–( a) S Focus V Directrix N P Axis V1 L L’ Length of Latus rectum : Eq. Of SLL’ : y = a P.O.I of this line and Parabola : x2 = 4a (a)

Standard Form of Parabola- Special Point / lines
Focus : ( 0,–a) , Vertex : ( 0,0) Axis : x = 0 , Directrix : y =( a) Length of Latus rectum : Eq. Of SLL’ : y = – a S Focus V Directrix N P Axis V1 L L’ P.O.I of this line and Parabola : x2 = –4a (–a)

Algorithm to Find out special points - Standard Form
Vertex : (0,0) Axis : Put Second degree variable = 0 Focus : If second degree variable is y : (  a,0) If second degree variable is x : (0,  a) Line of Directrix : If second degree variable is y : x = – (  a) If second degree variable is x : y = – ( a) Length of Latus rectum : 4a

Class Exercise Find the focus, line of directrix and length of latus rectum for the parabola represented by Solution : Axis : Put Second degree variable = 0 x = 0 Focus : If second degree variable is x : (0,  a) Line of Directrix : If second degree variable is x : y = – ( a) Length of Latus rectum : 18 units

Class Exercise For what point of parabola y2 = 18 x is the y-coordinate equal to three times the x-coordinate? Solution : As this point is on parabola

General Form - Parabola
Focus : (x1,y1) , Line of directrix : Ax + By + 1 = 0 Let P be (  , ) e =1

General Form - Parabola

General Form - Parabola
One of the Condition for second degree equation to represent parabola

Class Test

Class Exercise Solution : Pre – session - 6

Class Exercise Solution :
If the focus is (4, 5) and line of directrix is x + 2y + 1 = 0, the equation of the parabola will be ? Solution :

Class Exercise Solution :
If the focus is (4, 5) and line of directrix is x + 2y + 1 = 0, the equation of the parabola will be ? Solution :

General Form - Parabola
can be converted in to

Algorithm to find Special points/ lines - General Form
Convert the given equation in to general form e.g. : y2 – 6y + 24x – 63 = 0 Can be written as : y2 – 6y + 9= – 24x + 72 Transform the same in to Standard form

Algorithm to find Special points/ lines - General Form
Find special points/ Line in transformed axis ( X, Y) Vertex : (0,0), Axis : Y = 0 Focus : (– 6,0) ( as of form y2 = 4ax ) , Directrix : X = – (– 6) or X = 6 Reconvert the result in to original axis ( x,y) Vertex : X = 0  x – 3 = 0  x = 3 Y = 0  y – 3 = 0  y = 3 ( 3 ,3) Focus : ( –3 , 3) , Directrix : x = 9

Class Exercise Solution : Transform in to Standard form
(0, –4); x = – (b) (–4, –2); x = –2 (c) (–2, –4); y = – (d) (0, –4); x = –4 Solution : Transform in to Standard form Find special points/ Line in transformed axis ( X, Y) Focus - ( 2,0) ; Line of Directrix : X = –2

Class Exercise Solution :
(0, –4); x = –2 (b) (–4, –2); x = –2 (c) (–2, –4); y = –4 (d) (0, –4); x = –4 Focus - ( 2,0) ; Line of Directrix : X = –2 Solution : Reconvert the result in to original axis ( x,y) Focus – ( 0 , –4) Practice Exercise - 9

Class Exercise In a parabola , vertex is at (1,1) and line of directrix is x + y = 0. Equation of parabola ? Solution : Axis is y – x = k Vertex lies on the axis Axis : y – x = 0 P.O.I of axis and Directrix : (0 , 0) Let focus be ( h, k) Focus – (2 ,2)

Class Exercise In a parabola , vertex is at (1,1) and line of directrix is x + y = 0. Equation of parabola ? Solution : Focus – (2 ,2) ; Line of directrix : x+y = 0

Class Exercise Draw the rough shape of the curve represented by y=ax2+bx+c; where b2– 4ac > 0 , > 0 and b < 0 and find out vertex and axis of parabola. Compare the results with solution of ax2+bx+c = 0 when b2– 4ac > 0 and a > 0 Transforming the given equation to general form, we get

Class Exercise Transforming the equation into standard form, we get
Shape is parabola

Class Exercise Axis: X = 0 and a > 0, b < 0, D > 0,

Class Exercise y=ax2+bx+c and a > 0, b < 0, D > 0,
ax2 + bx + c = 0 (i.e.y = 0) for two real values of x . (  ,  )

Thank you