1. Mathematics

2. PARABOLA - SESSION 1

3. 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

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

5. 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

6. 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

7. 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

8. 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

9. 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.

10. Class Exercise

11. 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

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

13. 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

14. 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

15. 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

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

17. 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’

18. 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)

19. 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)

20. 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)

21. 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

22. 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

23. 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

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

25. General Form - Parabola

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

27. Class Test

28. Class Exercise
Solution :
Pre – session - 6

29. 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 :

30. 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 :

31. General Form - Parabola
can be converted in to

32. 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

33. 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

34. 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

35. 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

36. 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)

37. 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

38. 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

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

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

41. 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 . (  ,  )

42. Thank you