# 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)–  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. 