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 + 2xy + 3y2 + x – y + 1 = 0 represents an
ellipse.
Soln: Given x2 + 2xy + 3y2 + x – y + 1 = 0 –––– (1)
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ––––– (2)
Comparing, we get
a = 1 2h = 2 b = 3
h =1
h2 – ab = (1)2 – 1(3) = 1 – 3 = -2 < 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.
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