mathematics. parabola - session 1 session objectives definition of conic section eccentricity...

Mathematics

PARABOLA - SESSION 1

Session Objectives

• Definition of Conic Section

• Eccentricity

• Definition of Special Points

• Condition for Second degree equation to represent different conic sections

• Standard Form of parabola

• General Form of parabola

• Algorithm for finding special points/ lines

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

Ellipse



Ellipse

• If plane is not perpendicular to the axis

• Does not pass through base

Parabola



Parabola

• If plane is parallel to the generator

Hyperbola



Hyperbola

• Two similar cones

• Plane parallel to the axis

Class Exercise

Are the following be a conic section?

If yes, how they can be generated by intersection of cone(s) and plane.

(i) Point

(ii) Pair of straight lines

Class Exercise

Locus Definition

Locus Definition

Locus of a point moves such that

• Ratio of its distance

• Ratio - Eccentricity

• Fixed Point - Focus

• Fixed Line - Line of Directrix

NPPS

EccentricityPN

Fixed Line

S

Fixed Point

• from a fixed point

• & from a fixed line is constant

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

Directrix

NP

S

Focus

Axis

Vertex :

Meeting point of Curve and axis

Vertex

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

NP

S

Focus

AxisVertex

Standard Form of Parabola

e =1

• Axis is x- axis , y = 0

• Vertex - ( 0,0)

• Focus - ( a,0)

Directrix

NP

S

Focus

AxisVertex

V

As e = 1 , SV = VV1

V1

Equation of Directrix : x a

2 2PS a 0

Let P be ( , )

2

aPN a

1 0

Standard Form of Parabola

e =1

Directrix

NP

S

Focus

AxisVertex

VV1

PSNow e 1

PN

2 2PS PN a a

2 22a a

2 4a

2Equation of parabola y 4ax

Standard Form of Parabola- Special Point / lines

Directrix

NP

S

Focus

AxisVertex

VV1

2Equation of parabola y 4ax Focus : ( a,0) , Vertex : ( 0,0)

Axis : y = 0 , Directrix : x = – a

P.O.I of this line and Parabola :

y2 = 4a (a)

L

L’

Length of Latus rectum :

Eq. Of SLL’ : x = a

L a,2a L ' a, 2a

y 2a

LL ' 4a


2Equation of parabola y 4ax Focus : ( -a,0) , Vertex : ( 0,0)

Axis : y = 0 , Directrix : x =–(– a)


y2 = – 4a (–a)

N

S

Focus

Vertex

V

Directrix

P

Axis V1

L

L’


Eq. Of SLL’ : x = –a

L a,2a L ' a, 2a

y 2a

LL ' 4a


2Equation of parabola x 4ay

Focus : ( 0,a) , Vertex : ( 0,0)

Axis : x = 0 , Directrix : y =–( a)


x2 = 4a (a)


Eq. Of SLL’ : y = a

L 2a,a L ' 2a,a

x 2a

LL ' 4a

S

Focus

V

Directrix N

P

Axis

V1

LL’


2Equation of parabola x 4ay

Focus : ( 0,–a) , Vertex : ( 0,0)

Axis : x = 0 , Directrix : y =( a)


x2 = –4a (–a)


Eq. Of SLL’ : y = – a

L 2a, a L ' 2a, a

x 2a

LL ' 4a

S

Focus

V

Directrix N

P

Axis

V1

LL’

Algorithm to Find out special points - Standard Form

2 2y 4ax , x 4ay

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

Axis : Put Second degree variable = 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

Find the focus, line of directrix and length of latus rectum for the parabola represented by

2x –18y.

Solution : x = 0

418

0 ,

29

y

29

0 ,

29y

Class Exercise

For what point of parabola y2 = 18 x is the y-coordinate equal to three times the x-coordinate?

Solution :

3,beintpotheLet

As this point is on parabola 183 2

02 , 0062 ,,beintPo

General Form - Parabola

Focus : (x1,y1) ,

Line of directrix : Ax + By + 1 = 0

e =1

2 21 1Dis tance from Focus x y

Let P be ( , )

2 2

A B 1Dis tance from Directrix

A B

2 21 1 2 2

A B 1x y

A B

22 21 1 2 2

A B 1x y

A B


22 21 1 2 2

A B 1x y

A B

2 2 2 2 2 2 2 21 1

2 2 2 21 1

B A 2AB 2 x A B A 2 y A B B

x y A B 1

2 2 2 2B x A y 2ABxy ........


Comparing with

2 2ax by 2hxy 2gx 2fy c 0

2 2a B , b A , h AB

2h ab

2 2 2 2B x A y 2ABxy ........

One of the Condition for second degree equation to represent parabola

Class Test

Class Exercise

?abhthatenoughitis,parabolarepresent

tocfygxhxybyaxFor

2

22 0222

Solution : 02 222 chbgaffghabc

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 ?

Let P ,β be the point whose locus be the

desired parabola

2 2Dis tance from Focus 4 5

2 2

2 1Dis tance from Directrix

1 2

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 ?

2 2

2 2

4 5e

2 1

1 2

2 2 2

5 4 5 2 1

2 24 4 42 54 204 0

2 24x y 4xy 42x 54y 204 0


2

2

y k x Or

x k y

2I f h ab2 2ax by 2hxy 2gx 2fy c 0

can be converted in to

Algorithm to find Special points/ lines - General Form

2

2

y k x Or

x k y

1. Convert the given equation in to general form

e.g. : y2 – 6y + 24x – 63 = 0

Can be written as : y2 – 6y + 9= – 24x + 72

2y 3 24 x 3

2. Transform the same in to Standard form

2Y 24X , where Y y 3 and X x 3

Algorithm to find Special points/ lines - General Form

3. Find special points/ Line in transformed axis ( X, Y)

4. Reconvert the result in to original axis ( x,y)

2Y 24X , where Y y 3 and X x 3

2For Y 24X ,

Vertex : (0,0), Axis : Y = 0

Focus : (– 6,0) ( as of form y2 = 4ax ) ,

Directrix : X = – (– 6) or X = 6

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

2

The focus and the directrix for

the parabola represented by

y 4 8 x 2 is

(a) (0, –4); x = –2 (b) (–4, –2); x = –2(c) (–2, –4); y = –4 (d) (0, –4); x = –4

Transform in to Standard formSolution :

2Y , where Y y 4 and XX x8 3

Find special points/ Line in transformed axis ( X, Y)

Focus - ( 2,0) ; Line of Directrix : X = –2

Class Exercise

2

The focus and the directrix for

the parabola represented by

y 4 8 x 2 is

(a) (0, –4); x = –2 (b) (–4, –2); x = –2(c) (–2, –4); y = –4 (d) (0, –4); x = –4

Solution : Focus - ( 2,0) ; Line of Directrix : X = –2

Reconvert the result in to original axis ( x,y)

For Focus, X 2 2 2 xx 0

Y 0 y y4 0 4 Focus – ( 0 , –4)

Line of directrix is x 2 2 xX 2 4

Practice Exercise - 9

Class Exercise

Solution :

In a parabola , vertex is at (1,1) and line of directrix is x + y = 0. Equation of parabola ?

Axis is y – x = k

Vertex lies on the axis k 1 1 k 0

Axis : y – x = 0

P.O.I of axis and Directrix : (0 , 0)

Let focus be ( h, k) 0 h 0 k

1 h 2 ; 1 k 22 2

Focus – (2 ,2) Vertex is mid point between focus and

P.O.I of axis and directrix

Class Exercise

Solution :

In a parabola , vertex is at (1,1) and line of directrix is x + y = 0. Equation of parabola ?

Focus – (2 ,2) ; Line of directrix : x+y = 0

2 2

Let P , be the point whose locus isdesired parabola

2 21

2

2 2 2 8 8 16 0

2 2Eqn. of parabola : x y 2xy 8x 8y 16 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

2y – c bx x

a a

Class Exercise

2y – c bx x

a a

2 22

2 2

b b y c bx 2 x –

2a a a4a 4a

2 2b 1 b – 4acx y

2a a 4a

Transforming the equation into standard form, we get

2 1 bX Y, where X x

a 2a

Shape is parabola

DY y

4a

Class Exercise

2 1 bX Y, where X x

a 2a

–bVertex X coordinate 0 x

2a

Axis: X = 0b

x –2a

b b Dx – Vertex – , –

2a 2a 4a

and a > 0, b < 0, D > 0,

y

x

x = – —b2a

— , b2a

– — D4a

–O

DY y

4a

DY coordinate 0 y

4a

Class Exercise

y=ax2+bx+c

b b Dx – Vertex – , –

2a 2a 4a

and a > 0, b < 0, D > 0,

2 1X Y

a

ax2 + bx + c = 0 (i.e.y = 0) for

two real values of x . ( , )

2ax bx c 0 for x

y

x

x = – —b2a

— , b2a

– — D4a

–O

2ax bx c 0 f or xor x

Thank you

mathematics. parabola - session 1 session objectives definition of conic section eccentricity...

Documents

x axis

degree variable

astandard form of parabola

special point lines

plane parabolaif plane

plane circleif plane

plane ellipseif plane

parabola session