1.3 CONICS

Conic: A conic is defined as the locus of a point which moves such that its distance from a fixed point is always ‘e’

times its distance from a fixed straight line.

Focus: The fixed point is called the focus of the conic.

Directrix: The fixed straight line is called the directrix of the conic.

Eccentricity: The constant ratio is called the eccentricity of the conic.

General equation of a conic  ax²  +  2hxy  + by² + 2gx + 2fy + c = 0  represents

(i) a circle if a = b and h = 0.

(ii) a parabola if  h² = ab.

(iii) an ellipse if h² < ab.

(iv) a hyperbola if h²> ab.

PART – A

Example  : Prove that the equation  x²  +  6xy  + 9y² + 4x + 12y – 5 = 0 is a parabola.                                                       

Soln:    x²  +  6xy  + 9y² + 4x + 12y – 5 = 0                    —————–   ( 1 )

            Condition for  ( 1 ) to represent parabola is  h² = ab

            From ( 1 )   a =  1,    b = 9

            2h = 6  ⇒  h = 3

            h² = ab

            3²  =   1 ( 9)

            9 = 9.                 ∴  ( 1 )   represents a parabola.

Example  :  Show that the equation  x²  + 4y² – 4x +24y + 31 = 0  represents an  

                      ellipse.

Soln:  Given    x²  + 4y² – 4x +24y + 31 = 0   –––– (1)

ax²  +  2hxy  + by² + 2gx + 2fy + c = 0    ––––– (2)

                   Comparing, we get

                        a = 1                2h = 0             b = 4               

                                                   h =0

                        h² –  ab =  (0)²  –  1(4)  =  0 – 4 =  -4 < 0                                     

                        Given equation ( 1 ) represents ellipse.

General Equation of a Conic

‘S’  denotes Focus                                                  

Line XM denotes Directrix

SP / PM  = e                                                                          

Note:

(i) If  e< 1,  the conic is called an ellipse.

(ii) If  e = 1,  the conic is called aparabola.

(iii) If e>1, the conic is called a hyperbola.

Parabola with vertex (0,0), focus- S (a,0), latus rectum and equation of directrix

y² = 4ax is the equation of the parabola

Types of Parabolas with vertex (h,k), focus, latus rectum and equation of directrix

Part – C

1.    Find axis, vertex, focus and equation of directrix for y² + 8x – 6y + 1 = 0

Soln:   y² – 6y = – 8x – 1 

           y² – 6y + 9 = – 8x – 1 + 9   ( Adding 9 on bothsides)

          (y – 3)² = – 8x + 8

                        = – 8(x – 1)

          (y – 3)² =    – 8(x – 1)

           This is of the form  Y²  =  – 8X                               (  y²  =  -4ax)  (open leftward)

                         Where  Y = y – 3   and   X  =  x – 1

           4a  = 8

              a =  2 

             vertex  (0 , 0) for X , Y

           Y = 0  ⇒   y – 3   = 0

                                  y = 3

             X  = 0    ⇒   x – 1 = 0

                                x = 1

                    The vertex is ( 1, 3)

        Focus: 

           The focus (X = – a,Y = 0)

              X  = -a

              x  – 1  =  – 2

              x  = – 1

              Y = 0  ⇒   y – 3   = 0

                                  y = 3

                     Focus is ( -1, 3 )

           Equation of directrix is  X – a = 0.

                                                   i.e., X – 2 = 0 

                                                 ⇒ x – 1 – 2 = 0

                                                  ⇒ x – 3 = 0 

               Latus rectum X + a = 0

                                i.e., x – 1 + 2 = 0 x + 1 = 0 

             Length of latus rectum = 4a = 4(2) = 8      

2) Find vertex, focus , equation of directrix  and latus rectum for   x² – 4x – 5y – 1 = 0

Soln:    x² – 4x – 5y – 1 = 0

             x² – 4x = 5y + 1 

             x² – 4x + 4 = 5y + 1 + 4   ( Adding 4 on bothsides)

            (x – 2)² = 5y + 5

                           = 5(y + 1)

          (x – 2)²  = 5(y + 1)

           This is of the form  X²  =  4aY                              

                        Where  X = x – 2   and   Y  =  y + 1

           4a  = 5

              a =  5/4 

             Therefore  1.  The vertex (X = 0, Y = 0)

                                          X  = 0    ⇒   x – 2 = 0  ⇒   x = 2

                                          Y  = 0   ⇒  y + 1 = 0  ⇒   y = – 1

                                               The vertex is ( 2, – 1)

                                    2.  The focus (X = 0, Y = a)

                                           is (x = 2, y + 1 = 5/4) = (2, 1/4)

                                    3.  The directrix equation is Y = -a  or  y + 1 = -5/4  or  4y + 9 = 0

                                    4.  The latus rectum is 4a = 4(5/4) = 5.