Parabola:

Definition: The set of all points whose distance from fixed point and fixed line is same.

Equation Of Parabola:

When  Equation of parabola parallel to y-axis

OR

Symmetric about y-axis

OR

Y=0

Then

                       (x-h)²=4a(y-k)

When  Equation of parabola parallel to x-axis

OR

Symmetric about x-axis

OR

x=0

Then

                        (y-k)²=4a(x-h)

   

       Exercise# 12.2

Q#1:
  1)    Y²= 6x
Solution:  Y²=4(6/4)x       By multiplying on right side “4/4”

  Focal length= a = 6/4 = 3/2  

Vertex(h,k)=(0,0) 

 Focus=(3/2 , 0)

Directrix line x= -3/2    2)    Y²= -10x

Solution: Y²= -4(10/4)x    By multiplying on right side “4/4”

 Focal length= a = 10/4 = 5/2 

Vertex (h,k)= (0,0)

 Focus = (-5/2 , 0)

 Directrix line x= 5/2
    3)     , 4) , 5) are same to question # 1), 2);

     6) x² – 40y= 0  
 Solution:    
X² – 40y = 0  

 X² = 40y    

X²  = 4(10)y  
  So;    

Focal length= a= 10    

Vertex(h,k)=(0,0)   

Focus = (0,10)   

 Directrix line    y = -10

 (y-3)² = 6(x-2)  
  Solution:    

 (y-3)²= 6(x-2)   

  (y-3)² = 4(6/4)(x-2)     

Focal length = a = 6/4 = 3/2     

Vertex(h,k) = (2,3)     

 Focus = (2+3/2, 3)                = (7/2, 3)     

 Directrix line   x = 2-3/2= 1/2   

Q # 8,9,10 are same to Q # 7.

11) x² – 4x + 2y = 1   
Solution:   
X² – 4x+2y = 1  

X² – 4x = 1 – 2y 

(x)² – 2(x)(2) + (2)² – (2)² = 1 – 2y 
  
(x – 2)² – (2)² = 1 – 2y  

 (x – 2)² = 1- 2y + 4   

 (x – 2)² = 5 – 2y   

 (x – 2)² = 5(1 – 2/5y)    

(x – 2)² = 4(5/4)(1 – 2/5y) 

So;  

Focal length = a = 5/4  

Vertex(h ,k) = (2 , 2/5)   

Focus = (2 , 2/5+5/4) = (2 , 33/20)   

Directrix line   y = - 5/4 

Q # 12 – 16 are same to Q # 11.



In Exercises 17 – 13, find the Equation of Parabola satisfying the following conditions.


17) Vertex (0 , 0)  ,  focus (3 , 0),

Solution: 

From focus we can find focal length = a = 3; 

And from Vertex(h , k) = (0 , 0) 

And of we look at focus then we can write Equation of parabola as 

(y - k)² = 4a(x - h)              (A). 

Because change is in x- axis of focus So; by using vertex and focal length then equation (A) Becomes   

(y - 0)² = 4(3)(x - 0)   

Y² = 12x        Answer.          

Q # 18 is same to Q # 17.


19) Vertex (0 , 0) , directrix  x = 7.

 Solution: 

From directrix line  x = 7

We can findFocal length = a = 7

And

Focus = (-7 , 0)

And
 Vertex (h, k) = (0 , 0)
So equation of parabola is 
(y - k)² = - 4 a (x - h)         (A)

By using the above two points in equation (A)

(y - 0)²= - 4 (7)(x - 0)Y² = - 28 (x)Y² = -28x.


           Q # 20 is same to Q # 19.

Ellipse:

Defination :- The sum of all those points whose sum of distance from two fixed point is constant.

   In ellipse there are two types of axis

                a) Majore axis

                b) Minor axis

Equation of Ellipse when Major axis is along x-axis:

       (x-h)/a²  +(y-k)/b²  =  1

Equation of Ellipse when Minor axis is along x-axis:

       (y-k)/a²  +(x-h)/b²  =  1

Diagram of Ellipse:

•   Interesting Information 

We can find any equation from the following formulas:
If    e = c/a = 1 (Then Equation will be Parabola)
If    e = c/a < 1 (Then Equation will be Ellipse)
If    e = c/a > 1 (Then Equation will be Hyperbola)

Exercsie # 12.3

In Exercise 1-14, sketch the ellipse. Label the foci and the ends of the major and minor axis.

1)    x² / 16  +  y² / 9 = 1

Solution

  a² = 16  ;      b² = 9

So;

Major axis is along  x - axis

Therefor:

Focal length=  a = 4 ( major axis )

b = 3 ( with both “ - ” and “ + ” sign )

We know that

a² = b² + c²

16 = 9 + c²

c²  = 16 – 9

c² = 7

c =   7 ( square root of " 7 " with both sign " - " & " + ")             

foci = ( + 7 , 0) , ( -7 , 0)

Ends of  minor  axis = ( 0 , 3) , ( 0 , - 3)

Ends  of  major  axis = ( +4 , 0) , ( -4 , 0)

   

Written By M. Sulaiman