CIRCLES and PARABOLA

PARABOLA

Parabola is the locus of a point which moves so that its distance from a fixed point and a fixed line are equal.

The fixed point is called the focus of the parabola and the  line is called the directrix of the parabola.

      DD’  –  directrix

      E, E’  –  end points of the latus rectum

      V  –  vertex of the parabola

       F – focus of the parabola 

      VF = a   –  focal distance

       DD’V =  VF =  a 

      EF = FE’  =  2a

  The axis of the parabola is horizontal ( OX ).

From the above figure and by definition of parabola :

 PF  =  PQ  

Ö ( x – a )² + ( y – 0 )²    =  ( x + a ) 

Square both sides

( x – a )² + y²  = ( x + a )²

X² – 2ax + a²  + y²  =  x²  + 2ax  + a²

X² – x² + a² – a² + y² = 2ax + 2ax

y² =  4ax 

Standard equation of a parabola having horizontal axis and opens to the right.        

                                                          

From the figure above and by definition of parabola :

PF  =  PQ

Ö( x – 0)² + ( y – a )²   =  y + a

Square both sides

X² + y² – 2ay + a²  =    y² + 2ay + a²

X² + y² – y² + a² – a²  =  2ay  + 2ay

X² = 4ay

Standard equation of a parabola with vertical axis (OY) and opens upward.

General Equation of a Parabola

1.  x² + Dx + Ey + F = 0 ,  E ¹ 0,   with vertical axis ( OY )

2.  y² + Dx + Ey + F = 0,   D ¹ 0,  with horizontal axis (Ox )

Standard Equation of the Parabola

A. with vertex at V( 0, 0 )

      1.  y² =  4ax  ----   opens to the right

      2.  y² =  – 4ax  ----  opens to the left

      3.   x² = 4ay  ----  opens upward

      4.   x² =  – 4ay  ----  opens downward  

     

B. with vertex at  V( h, k )

      1.  ( y – k )² =  4a( x – h )    ----   opens to the right

      2.  ( y – k )² =  – 4a( x – h)  ----  opens to the left

      3.   ( x – h )² = 4a( y – k )   ----  opens upward

      4.  ( x – h )² =  – 4a( y – k   ----  opens downward  

Examples.

Draw the graph of the following parabola.

  1.  2x² – 8y = 0                         3.  x² + 2x – 6y = 2

  2.  3y² + 18x = 0                       4.  y² + 2y – 6x = 2 

 5.  x² + 4y – 4 = 0                      8.  y² + 8x  + 16 = 0

 6.  y² – 4x + 4 = 0                      9.  x² = 4x + 4y 

 7.  y² + 2x + 6y + 17 = 0          10.  x² – 2x + 2y + 7 = 0

QUADRATIC FUNCTIONS

        A function of the form y = ax² + bx + c,  where  a ¹ 0 is a quadratic function. Since the equation has the form of the general equation of a parabola  x² + Dx + Ey + F = 0, the graph of every quadratic function is a parabola with a vertical axis. It is concave upward if a is positive or concave downward if a is negative.

         At the vertex of the parabola,  x = – b/2a. This value of x gives the function its minimum value if a is positive or concave downward if a is negative. The equation of the axis of the parabola is x = – b/2a.   

        As x increases, many functions increase to a maximum value, decreasing thereafter, or decrease to a minimum and begin to increase. In such cases, it is a problem of prime importance to determine the maximum or minimum value. To solve problems for functions in general, differential calculus is required. For the quadratic function it may evidently be solve by merely finding the vertex of the parabola, since the ordinate of the vertex is the greatest or least ( algebraically ) of all the ordinates, according as the parabola opens downward or upward.

STEPS IN GRAPHING A QUADRATIC FUNCTION

  1. Compute x = – b/2a, and then the corresponding value  of y

      to obtain the coordinates of the vertex of the parabola.

       

  2. If the value of  x = – b/2a is convenient, form a table by using

      pairs of values of x where each pair, the values are

      equidistant from the vertex on either side. The value of y

      corresponding to any pair will be equal.

  3. If the value of  x = – b/2a  is inconvenient, form the table by

      choosing values of x arbitrarily on either side of  x = – b/2a.

Example.  Determine the critical points whether it is maximum or minimum.

                1.  y  =  x² – 10x + 25 

                2.  y =  4x –  x²                                               

                3.  y =  2x² – 10x + 3

                4.  y = 15x – 3x²  

                                                          

Application problems.

Use quadratic function to evaluate the following problems. 

 1. Find the dimensions of the largest rectangular lot that can be

     enclosed by 100 m of fencing material.

      

 2. A rectangular field is to be fenced off along the bank of a

   river. If no fence is needed along the river, what is the shape

  of the field requiring 400 m of fence if the area is to be

  maximum?

   1. Solution

       Let  x and y be the dimensions (length and width) of the lot       

            Area         :   A = xy

               Perimeter :    2x + 2y = 100                                      

                                x +  y = 50                            

                                y = 50 – x 

                            x(50 – x ) = A                                                                                             

                             50x – x²  =  A                                 

                           Standardize the equation :

                               X² – 50x = – A                                                                

                            x =  – b/2a  = – (–50) / 2 ( 1 )

                            x =  25 m  

                            y = 50 – x

                            y = 50 – 25

                            y = 25 m

                  Alternately :       

                                By completing the square :

                              x² – 50x + 625 = – A + 625  

                             ( x – 25 )² = – ( A – 625 )

                             x – 25 = 0        |   A – 625 = 0

                             x = 25 m          |   A = 625 m²   

                            y = 50 – 25

                              = 25 m

               The dimensions are 25 m by 25 m.  

               The maximum area is  625 m²       

           

  B.  Use quadratic functions to solve the problems. 

     1. A rectangular field is to be enclosed, and divided into three

        lots by fences parallel to one of the sides. Determine the

      dimensions of the largest field that can be enclosed by

      2,400 m of fencing material.

  

    2.  A rectangular lot  is to be fenced off along a highway. If the

         fence along the highway cost  P 150 per meter, and on the

         other sides  P 100 per meter, determine the dimensions of

         the largest lot that can be enclosed  for a budget of

         P 100,000.

  

   3. A rectangular field is to be fenced off along the bank of a

       river. If no fence is needed along the river, what is the shape

      of the  field requiring 400 m of fence if the area is to be

      maximum?

                                    

CIRCLE

A circle is a locus of a point which moves in a plane so that its distance from a fixed point remains constant. The fixed point is called the center of the circle and the constant distance is the radius of the circle.

            By the definition of a circle, we have  CP  =  r,  or

              

                        Ö ( x – h)² + ( y – k )²           =  r

           By squaring both sides of the equation, we have

                                ( x – h )² + ( y – k )²  =  r²  

      which is called the standard form of the equation of a circle.

         Expanding the equation we have,

                             x² – 2hx + h² + y² – 2ky + k²  =  r²,  or

                             x²  +  y²  – 2hx  - 2ky  + ( h² + k² – r² )  =  0. 

                        If  -2h = D ,  -2k = E ,  (  h² + k² – r² ) = F       

        Then this equation is of the form    

                             X² + y² + Dx  +  Ey  +  F  =  0 ,

         which is the general  form of the equation of a circle.

Given a general equation of a circle, we need to transform it into the standard form so the
center and the radius can be determined and the graph can be drawn.

                                    

Example.

   1.  Find the center and the radius of the circle  x² + y²  =  16.

                  Solution :

                                    x² + y²  =  16

                                    ( x – 0 )² + ( y – 0 )² = 4² 

                                    C ( 0, 0 ),   r = 4 

  2.  Find the center and the radius of the circle 

         x² + y² + 4x – 2y  =  4  and draw the graph.

                  Solution :

                        x² + y²  +  4x – 2y  =  4

                       x² + 4x  + 4  + y² – 2y + 1  = 4 + 4 + 1

                      ( x + 2 )²  +  ( y  – 1 )²  =  3²   

                        C ( – 2,  1 ) ,   r  =  3

Exercises :

Find the center and the radius of the following circle and draw the graph.

 1.  x²+y² + 4x – 6y–12  = 0          6.  x²+ y² – 10x + 9 = 0     

 2.  x² + y² – 4x = 5                       7.  x² + y²+ 10x –  6y = 2 

 3.  x² + y² + 10y – 11 = 0             8.  x² + y² – 16y +15 = 0   

 4.  x² + y² –10x + 6y = 2              9.  x²+ y² +10x – 4y –7 = 0  

 5.   x² + y² +14y + 24 = 0           10.  x²+ y² – 4x + 6y –12 = 0

    

CIRCLES DETERMINED BY THREE CONDITIONS

Find the equation of the circle that passes through the points (-1,2),

(1,1) and (2,3).

Two solutions are available for the problem

   a) First solution is by using the general equation.

   b) Second solution is by using standard equation.

       To find the center of the circle use the intersection of the  

       perpendicular  bisector, then use distance formula to find the  

       radius of the circle.

 Solution by the general equation

       at (-1,2) :   (-1)² + 2² – D  + 2E  + F  = 0

                                       5 – D + 2E + F = 0    ¬ EQ 1

      at (1,1) :    1²  +  1²  +  D  +  E  +  F  =  0    

                                    2  +  D  +  E  +  F  =  0  ¬ EQ 2

      at (2,3) :  2²  +  3²  + 2D + 3E + F = 0

                                 13 + 2D + 3E + F = 0   ¬ EQ 3 

                      EQ 2 :  2 + D  +  E +  F  =  0

                  +  EQ 1 :  5 – D + 2E + F = 0                 .     

                                   7    + 3E  + 2F = 0   ¬ EQ 4  

                    2(EQ 1 ) :  10 – 2D + 4E + 2F  = 0

                     +  EQ 3 :   13 + 2D + 3E + F = 0     .        

                                       23  + 7E  + 3F  = 0  ¬ EQ 5   

                           3( EQ 4 ) :    21   +  9E  + 6F  = 0 

                       – 2( EQ 5 ) :  – 46 – 14 E – 6F = 0   .       

                                              – 25 – 5E  =  0

                                                           E = – 5 

                            EQ 4 :   F = ½ [ – 7 – 3(– 5)]  =  4   

                            EQ 2 :  D = – 2 – E – F 

                                        D = –2 – (–5) – 4  =  – 1

        Therefore, the equation is  x² + y² – x – 5y + 4 = 0.

   b) Solution by using the standard equation

                      Perpendicular bisectors :

                     ( x – 1 )² + ( y – 1 )²  = ( x – 2 )²  + ( y – 3 )²

             x² – 2x + 1 + y² – 2y + 1  =  x² – 4x + 4 + y² – 6y + 9

                        – 2x + 4x – 2y + 6y = 13 – 2

                                  2x + 4y = 11  ¬ EQ 1

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

               x² + 2x + 1  + y² – 4y + 4 = x² – 2x + 1 + y² – 2y + 1  

                            2x + 2x – 4y + 2y = 2 – 5

                              4x – 2y  = – 3   ¬ EQ 2

                           EQ 1 :  2x + 4y = 11  

                     2( EQ 2) :   8x – 4y = – 6        .    

                                       10x  =  5

                                            X = ½

             EQ 2 :  y = ½ (4x + 3 ) = ½ [ 4( ½ ) + 3 ]  =  5/2  

                                        C( ½ , 5/2 )

                        r²  =  ( 1 – ½ )²  + ( 1 – 5/2 )²  =   5/2

          By the standard equation of a circle :

                      ( x – ½ )²  +  ( y – 5/2 )²  =  5/2

                        x² – x  + ¼  + y² – 5y  + 25/4  =  5/2

                        x² – x + y² – 5y  + ¼ + 25/4 – 5/2  = 0

                        x²  –  y²  –  x  –  5y  +  26/4 –  10/4  = 0   

                        x²  –  y²  –  x  –  5y  +  4  =  0  

                    

   Therefore the equation of the circle is  x² – y² – x – 5y + 4  = 0

   Application problems.

     Find the equation of the circle

      1.  with r = 5 and C( 2 , -5 ).

      2.  containing ( -4 , 3) and C( 4 , -2 ) as end points of diameter.

      3.  tangent to the x-axis, C( 3 , 2 )

      4.  having (-1, -2) and (3, 4) as end points of the diameter.

      5.  tangent to the y-axis with C( -4, 3 ).

      6.  passing through ( 2,3 ), (6,1) and (4,-3)

      7.  passing through ( -3 , 1 ),  ( 5 , -3) and ( -2 , 4 ).

      8.  center on the y-axis and passes through the origin and (4, 2).

      9.  tangent to the axes and  C( -5, -5).

    10.  tangent to the axes and  r = 4.   ( 4 answers )

    11.  r = 5, tangent to the line 3x + 4y = 24 at ( 2 , 9/2)

    12.  C( 3, 5) and containing  the origin.

    13.  Find the equation of the circle that passes through the points

            ( -1 , 2 ),  ( 1 , 1 )  and ( 3 , 2 ).

    14.  Find the equation of the circle that passes through the point

            ( 9 , 7 ) and is tangent to both the y-axis and the

            line  3x – 4y  =  24.

     15.  Tangent to the axes,  center third quadrant,  radius 5.

     16.  Find the equation of the circle with the points ( 9 , 7 ) and ( -3 , 3 ) as end points of diameter.     

     17.  Tangent to the axes,  center 4^th quadrant,  radius 4.

     18.  tangent to the x-axis with C( 4, -3 ).

     19.  C( –2, –3 ) and  r =  5

     20.  C( –4,  5 )  and passing through  (  –1,  1 ) 

RADICAL  AXIS 

     If  x² + y² + Dx + Ey  + F = 0   and  x² + y² + D’x + E’y + F’ = 0,

are the equations of two non-concentric circles, then the equation

  x² + y² +  Dx + Ey + F + k(x² + y² + D’x + E’y + F’ ) = 0,   ( EQ 1 )   

 represents a circle for all values of k except  k = –1. If the given circles intersect 
in two distinct points, all members of the family 

( EQ 1 ) will pass through these points. This is evident since any point (x₁, y₁) 
that satisfies the given equations will make the left side of (EQ1) read 0 + k0, 
which is zero for any value of k. If the given circles are tangent at a point,  the 
family ( E Q 1)  will be tangent  to them  at  the point  of tangency.

 Example.

   Find the equation of the circle that passes through the points of intersection of the circle  x² + y² = 2x,  x² + y² = 2y,  and contains the point  ( a )  ( 2, 3 )   and  ( b )  ( 2, 1 ).  

          By  EQ 1 :     x² + y² – 2x  +  k( x² + y² – 2y ) =  0

          at (2 , 3) :     2²  + 3² – 2(2) + k [ 2² + 3² – 2(3)] = 0

                                 4 + 9 – 4 + k [ 4 + 9 – 6 ] = 0

                                              9 + k ( 7 ) = 0

                                                    k =  – 9/7

       The member of this family that passes through ( 2, 3 ) is

                          x² + y² – 2x  – (9/7)( x² + y² – 2y ) =  0

                        7x² + 7y² – 14x  – 9x² – 9y² +  18y =  0

                          – 2x² – 2y²  – 14x + 18y = 0

                               x²  + y²  +  7x – 9y  =  0

     

  In  EQ 1,  if we set  k = – 1, the linear equation

                    ( D – D’) x  +  ( E – E’ )y + ( F – F’) = 0  is obtained.

     This equation is called the radical axis of the two circles and has the following properties :

      1.  If two circles intersect in two distinct  points, their radical axis  

           contains the common chord of the circles.

      2.  If two circles are tangent, their radical axis is the common tangent to the circles at their point of tangency.

      3.  The radical axis of two circles is perpendicular to their line of

           centers.

      4.  All tangents drawn to two circles from a point on their radical axis have the same length.

  Find the radical axis of the following circles.

    1.  x²  + y²  +  4x – 8y  = 5  and  x²  + y²  +  10x – 6y  = 2

    2.  x²  + y²  – 4x + 8y  = 16  and  x²  + y²  – 10x + 6y + 9  =  0

    3.  x²  + y²  +  4x  = 12  and  x²  + y² – 6y  = 7

    4.  x²  + y²  – 8y  = 9  and  x²  + y²  +  10x  =  0

    5.  x²  + y² + 8y  =  9  and  x²  + y²  +  10x – 6y  =  2