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Mathematics

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PARABOLA - SESSION 1

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**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

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**Definition of Conic section**

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

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**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

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**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

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**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

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**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

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**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.

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Class Exercise

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**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

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**Eccentricity and Shapes of Conic Section**

e = 1 : Parabola e < 1 : Ellipse e = 0 : Circle e > 1 : Hyperbola

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**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

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**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

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**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

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**Standard Form of Parabola**

e =1 Directrix N P S Focus Axis Vertex V V1

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**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’

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**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)

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**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)

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**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)

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**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

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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

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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

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**General Form - Parabola**

Focus : (x1,y1) , Line of directrix : Ax + By + 1 = 0 Let P be ( , ) e =1

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**General Form - Parabola**

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**General Form - Parabola**

One of the Condition for second degree equation to represent parabola

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Class Test

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Class Exercise Solution : Pre – session - 6

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**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 :

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**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 :

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**General Form - Parabola**

can be converted in to

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**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

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**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

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**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

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**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

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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)

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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

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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

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**Class Exercise Transforming the equation into standard form, we get**

Shape is parabola

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

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**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 . ( , )

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Thank you

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