## 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  +  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)–  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.

## 1.2 ANALYTICAL GEOMETRY II

EQUATION OF CIRCLE

Definition:

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

Equation of the circle with centre (h, k)  and radius ‘r’ units.

CP = r

√(( x – h )2 + (y – k )2   =  r (Using distance formula)

(x – h )2 + (y – k )2    =   r2

Note:

The equation of the circle with centre (0, 0 ) and radius ‘r’ units is  x+ y2  = r2

#### Part – A

Example  :      Find the equation of the circle with centre (- 5, 7 ) and radius 5 units.

Soln:   We know that  the  equation of circle is (x – h )2 + (y – k )2    =   r2

Here h =  –  5,  k = 7  (given)   and   r  =  5

.     .        (x + 5 )2  + (y + 7 )2 =

x2 + 10x + 25  +  y2 +  14y + 49 = 25

x2  +   y2   + 10x  – 14y + 25 +49-25 = 0

x2  +   y2   + 10x  – 14y + 49 = 0

Therefore the equation of circle is x2  +   y2 + 10x  – 14y + 49 = 0

General equation of the circle:     x2  +   y2  + 2gx  + 2fy  + c = 0

Centre = (-g , -f )         and    radius     r =  √( g2  + f2 – c)

#### Part – A

Example  :   Find the centre and radius of the circle  x2  +   y2 +  2x  +  2y  –  7 = 0 .

Soln:    Given   x2 +  y2  +  2x  +  2y  –  7 = 0 .

We know the equation of circle is  x2  +   y2  + 2gx  + 2fy  + c = 0

2g  = 2                  2f = 2                  c = -7

g  =  1                      f =  1

centre = ( -g , -f )                              r = √( g2  + f2 – c)

=   ( – 1 , – 1 )                           r = √( (1)2  + (1)2 + 7)

r = √9                r =  3

#### Part – B

Example  :   Find the equation of the circle passing through the point  A (2, – 3) and having its centre at

C ( – 5 , 1).

Soln:                r =  √(( – 5 – 2 )2 + (1 + 3 )2)

=  √( (- 7)2  + (4)2 )

=  √ ( 49 + 16)

r    =  √65

We know that  the  equation of circle is (x – h )2 + (y – k )2    =   r2

Here h = -5,  k = 1  (given)   and   r  =  √65

(x + 5 )2  + (y – 1 )2 =  r2

x2   + 10x + 25  +  y2  –  2y + 1 = 65

x2  +   y2  +  10x  – 2y + 26 – 65 = 0

x2  +   y2  + 10x  –  2y  – 39 = 0

Therefore the equation of circle is   x2  +   y2  + 10x  –  2y  – 39 = 0

Equation of  the Tangent to a circle at the point (x1,y1) on the circle

Equation of the tangent to a circle x2  +   y2  + 2gx  + 2fy  + c = 0  at ( x1, y1)  is

x x1 + y y1 + g ( x + x1 ) + f ( y + y1) + c = 0

#### Part – B

Example  :   Find the equation of the tangent at (4,1) to the circle

x2  +   y2 –  8x  –  6y  + 21 = 0 .

Soln:    W.K.T  the equation of tangent at  ( x1, y1)  is

x x1 + y y1 + g ( x + x1 ) + f ( y + y1) + c = 0

Given   x2  +   y2  –  8x  –  6y  + 21 = 0

We know the equation of circle is  x2  +   y2  + 2gx  + 2fy  + c = 0

2g  = – 8                  2f = – 6                  c = 21

g  =  – 4                      f = – 3

x x1 + y y1 + g ( x + x1 ) + f ( y + y1) + c = 0

x (4) + y ( 1) – 4 ( x + 4)  – 3 ( y + 1)  + 21 = 0

4x  + y   – 4x – 16  – 3y – 3  + 21 = 0

– 2y + 2  = 0

∴  Equation of tangent is  y  – 1 = 0.

FAMILY OF CIRCLES

Concentric Circles:

Two or more circles having the same centre but differ in radii are called concentric circles.

Note:   Equation of concentric circles differ only by the constant term.

Example 1 :Show that the circles x2  +   y2 – 4x  + 2y +5 = 0  and  x2  +   y2 – 4x  + 2y – 8 = 0

are concentric circles.

Soln:    From the two given equations of the circles, we observe that the constant term alone differs

∴The given circles are concentric circles.

#### Part – A

Example 2 : Find the equation of the circle passing through the point (5 , 4 ) and concentric to the circle

x2  +   y2  –  8x  +  12y  +  15 = 0 .

Soln:    Equation of concentric circle be  x2  +   y2  –  8x  +  12y  + k = 0  ………….. ( 1 )

Put  x = 5,  y = 4  in  ( 1 )

(5)2 +  (4)2   – 8 ( 5 )  +  12 ( 4 ) + k = 0.

25  +  16 – 40 + 48 + k = 0.

89 – 40 + k = 0.

49 + k = 0

K  =  – 49

The required equation of the circle is  x2  +   y2  –  8x  +  12y  – 49 = 0

Orthogonal Circles:

Two circles   x2  +   y2  + 2g1x  +2f1y+c1 = 0 and  x2  +   y2  + 2g2x  + 2f2y +  c2 = 0 are said to be Orthogonal Circles if

2g1g2  + 2 f1 f2  =  c1 + c2

#### Part – B

Example :Prove that the circles  x2  +   y2  – 4x  + 6y + 4 = 0  and

x2  +   y2  + 2x  + 4y + 4 = 0 cut orthogonally.

Soln:    Given  x2  +   y2  – 4x  + 6y + 4 = 0   ––––– (1)      and

x2  +   y2  + 2x  + 4y + 4 = 0   ––––– (2)

From ( 1 )

2 g1  = – 4                   2 f1 = 6                c1 = 4

g1=  – 2                       f1= 3

From ( 2 )

2 g2  =2                    2 f2 = 4               c2 = 4

g2=  1                        f2= 2

The condition for orthogonally is

2g1g2  + 2 f1 f2  =  c1 + c

2(-2)(1) + 2(3)(      4+ 4

-4   + 12        =      8

8                      =        8

∴ The circles cut orthogonally.

### Contact of Circles:

Case ( i ) :

Two circles touch externally if the distance between their centres is equal to sum of their radii.

i.e  C1C=  r1  +  r2

Co-ordinates of point of contact are

P = ( (r1 x2 + r2 x1 ) / (r1  +  r2 ) ,  (r1 y2 + r2 y1 ) / (r1  +  r2 ) )

where  C1 = (x1,  y1)   and   C2  =  (x2,  y2)

Case ( ii ) :

Two circles touch internally if the distance between their centres is equal to difference of their radii.

i.e  C1C= r1  –  ror   r2  –  r1

Co-ordinates of point of contact are

P = ( (r1 x2 – r2 x1 ) / (r1  –  r2 ) ,  (r1 y2 – r2 y1 ) / (r1  –  r2 ) )

where  C1 = (x1,  y1)   and   C2  =  (x2,  y2)

#### Part – C

1.   Prove that the circles  x2  +   y2  + 2x  –  4y  – 3 = 0 and x2  +   y– 8x  +  6y  +7 = 0 touch each  other.

Soln:    Given  x2  +   y2  + 2x  –  4y  – 3 = 0 ––––– (1)      and

x2  +   y– 8x  +  6y  +7 = 0  ––––– (2)

From ( 1 )

2g1  = 2                    2f1 = – 4                c1 = -3

g1 =  1                       f1= – 2

centre  is  C1  = ( – g1 , – f1 )                 r1 = √( g12  + f12 – c)

=   ( -1 , 2 )                 r= √( (1)2  + (-2)2 + 3)

r= √( 1  + 4  +3 )

r1 = √8 = 2√2

From ( 2  )

2g2   = -8                    2f2 = 6                c2 = 7

g2  =  – 4                        f2 = 3

centre  is  C2  = ( – g2 , – f2 )                 r2 = √( g22  + f22 – c)

=   ( 4  , -3 )                  r= √( (-4)2  + (3)2  – 7)

r= √( 16  + 9  – 7 )

r2 = √18 =   √(9 × 2)  =   3√2

C1C2   =  √(( – 1 – 4 )2 + (2 + 3 )2)

=   √( (- 5)2  + (5)2 )

=   √( 25  + 25)

= √50 =   √(25 × 2)   =   5√2

C1C2   =   5√2

r1  +  r2  =   2√2  +   3√2

=   5√2  =   C1C2

∴The circles touch each other externally.

2. Prove that the circles  x2  +   y2   – 4x  +  6y – 112= 0 and  x2  +   y2   – 10x  – 6y + 14 = 0 touch  each  other.

Soln:    Given x2  +   y2  – 4x  +  6y – 112 = 0 ––––– (1)      and

x2  +   y2 – 10x  – 6y + 14 = 0  ––––– (2)

From ( 1 )

2g1  = -4                    2f1 = 6                c1 = -112

g1=  -2                       f1= 3

centre  is  C1  = ( – g1 , – f1 )                 r1 = √( g12  + f12 – c)

=   ( 2 , -3 )                        r= √( (-2)2  + (3)2 + 112)

r= √( 4  + 9  + 112 )

r1 = √125  = √(125 × 5)  =   5√5

From ( 2 )

2 g2  = – 10                    2 f2 = -6                c2 = 14

g2 =  -5                        f2 = – 3

centre  is  C2  = ( – g2 , – f2 )                 r2 = √( g22  + f22 – c)

=   ( 5 , 3 )                   r= √( (-5)2  + (-3)2  – 14)

r= √( 25  + 9  – 14 )

r2 = √20  = √(4 × 5)= 2√5

C1 C2   =  √(( 5 – 2 )2 + (3 + 3 )2)

=   √( (3)2  + (6)2 )

=    √( 9  + 36)

C1 C2   =   √45  =  √(9× 5)  = 3√5

r1  –  r2    =  5√5  –

=  3√5  =  C1 C2

∴The circles touch each other internally.

## Tamil Nadu Diploma Engineering Mathematics – II Unit – I ( 1.1 – Analytical Geometry – I) material 2020-21 ( N-Scheme)

UNIT – I     ANALYTICAL GEOMETRY

1.1    ANALYTICAL GEOMETRY  I

Straight Line:

When a variable point moves in accordance with a geometrical law, the point will trace some curve. This curve is known as the locus of the variable point.

If a relation in x and y represent a curve then

(i) The co-ordinates of every point on the curve will satisfy the relation.

(ii) Any point whose co-ordinates satisfy the relation will lie on the curve.

Straight line is a locus of a point.

Slope or gradient of a straight line:

The tangent of the angle of inclination of the straight line is called slope or gradient of the line.  If θ is the angle of inclination then slope = tan θ and is denoted by m.

i.e  m  =  tan θ

Equation of a straight line:

When ‘c’ is the y intercept and slope is ‘m’, the equation of the straight line is

y = mx+c

Formulae

1)    The length of the perpendicular distance from (x1,y1) to the line ax + by + c = 0 is

±  (ax1 + by1 + c)/ √(a2 + b2 )

2)    The distance between the parallel lines ax + by + c1 = 0  and  ax + by + c2 = 0  is

±  (c–  c2 )/ √(a2 + b2 )

ANGLE BETWEEN TWO STRAIGHT LINES:

Book Work:

Find the angle between the lines y = m1 x+c1  and y = m2x+c2.  Deduce the conditions for the lines to be

(i) parallel (ii) perpendicular.

Proof:

Let θ1 be the angle of inclination of the line y = m1x+c1.

Slope of this line is m1 = tanθ1.

Let ‘θ’2 be the angle of inclination of the line y=m2x+c2 .

Slope of this line is m2 = tanθ2 .

Let ‘θ’ be the angle between the two lines, then θ1 = θ2 + θ

Θ = θ1 – θ2

∴tan θ = tan (θ12)

=       (tan θ1- tan θ2)/(1 + tan θ1 tan θ2 )

tan θ   =  (m1 – m2)/(1 + m1 m2)

θ   =  tan^(-1)⁡((m1 – m2)/(1 + m1 m2))

(i) Condition for two lines to be parallel:

If the two lines are parallel then the  angle between  the two lines is zero

∴ tan θ = tan 0 = 0

(i.e)   =  0

m1  –    m2   =  0

∴   m=  m2

(ii)Condition for two lines to be perpendicular:

If the two lines are perpendicular then the angle between them

Θ =  900

∴ tan θ =  tan 900  =  ∞ =  1/0

(m1 – m2)/(1 + m1 m2) = 1/0

1 +  m1 m2  =  0

m1 m2  =  – 1

∴   For perpendicular lines, product of the slopes will be -1

Note:     1)  Any line parallel to the line ax+by+c = 0

will be of the form ax+by+ k = 0 (differ only constant term)

2) Any line perpendicular to the line ax+by+c =0

will be of the form bx – ay  + k = 0

WORKED EXAMPLES

Part – A

.    1.  Find the perpendicular distance from the point (2,3) to the straight line 2x+y+3=0.

Soln:    W.K.T  The length of the perpendicular distance from (x1,y1) to the line ax + by + c = 0  is

±  (ax1 + by1 + c)/ √(a2 + b2 )

Given straight line is    2x+y+3=0

Given point (x1,y1)  =  (2,3)

i.e   2(2 ) + 1(3) + 3/ √(22 + 12 )   =   10 / √5

2.        Find the distance between the line 2x+3y+4 = 0 and 2x+3y -1 = 0

Soln:   W.K.T  the  distance between the parallel lines

ax + by + c1 = 0  and  ax + by + c2 = 0  is

±  (c–  c2 )/  √(a2 + b2 )

Here  a = 2,  b = 3  ,   c1 =  4   and   c2 =  – 1

i.e   4 + 1/√(22 + 32 )     =   5 /√13

3.        Show that the lines 6x+y-11 =0 and 12x+2y+14 =0 are parallel

Soln:                6x+y-11 = 0                                                                       (1)

12x+2y+14 = 0                                                                  (2)

Slope of the line (1) = m1 =  – a/b

=  – 6/1

m1 =  – 6

Slope of the line (2) = m2 =  – a/b

=  – 12/2

=   – 6

m=  m2

∴  The lines are parallel.

4.           Find ‘p’ such that the lines 7x-4y+13 = 0 and px = 4y+6 are parallel.

Soln:                7x-4y+13 = 0                                                                       (1)

px – 4y+ 6 = 0                                                                 (2)

Slope of the line (1) = m1 =  – 7/-4

m1 =  7/4

Slope of the line (2) = m2 =  – p/-4

m2  =   p/4

Since (1) and (2) are parallel lines

m=  m2

7/4  =   p/4

4p    =  28

∴        p   =  7

5.           Show that the lines 2x+3y-7 =0 and 3x- 2y+4 =0 are perpendicular.

Soln:                2x+3y-7 =0                                                                       (1)

3x- 2y+4 =0                                                                     (2)

Slope of the line (1) = m1 =  – a/b

m1  =  – 2/3

Slope of the line (2) = m2 =  – 3/-2

m2 =   3/2

m1  m2 =  (- 2/3 ) ×  (3/2)

=  – 1

∴  The lines (1) and (2) are perpendicular.

6.      Find the value of m if the lines 2x + my = 4  and  x + 5y – 6 = 0  are perpendicular.

Soln:                2x + my – 4 = 0                                                                        (1)

x + 5y – 6 = 0                                                                        (2)

Slope of the line (1) = m1 =  – 2/m

Slope of the line (2) = m2 =  – 1/5

Since (1) and (2) are  perpendicular

m1 m2  =  – 1

(- 2/m ) (- 1/5)  =   -1

( 2/5m )    =  – 1

– 5m     =  2

∴           m    = – 2/5

Part – B

1.    Find the angle between the lines y = √3 x  and  x- y = 0

Sol:      y = √3 x

√3 x  –  y = 0                                                                                          (1)

x- y = 0                                                                                             (2)

Slope of the line (1) = m1 =    – coefficient of  x /  coefficient of y

=   – √3   /  – 1

m1 =  √3

tan θ1  =  √3  ⇒        θ1  =  600

Slope of the line (2) = m2 =    – coefficient of  x /  coefficient of y

=   – 1  /  – 1

m2 =  1

tan θ2  = 1 ⇒         θ2  =  450

∴  Θ = θ1 – θ2

Θ = 600– 450

Θ =  150

2.    Find the angle between the lines 3x + 6y = 8  and  2x  = – y + 5

Sol:      3x + 6y = 8

3x + 6y  – 8 = 0                                                                                          (1)

2x + y – 5 = 0                                                                                              (2)

Slope of the line (1) = m1 =    – coefficient of  x /  coefficient of y

=   –  3 /  6

m1 =  – 1 / 2

Slope of the line (2) = m2 =    – coefficient of  x /  coefficient of y

=   – 2  / 1

m2 =  – 2

W.K.T          tan θ   =  (m1 – m2)/(1 + m1 m2)

tan θ   =  (1 / 2 + 2)/(1 + ( – 1/2 ) ( – 2 ))

tan θ   =  3/2/2

tan θ   =   3/4

3.     Find the equation of the straight line passing through (-1,4)  and

parallel to x + 2y = 3.

Soln:   Let the equation of line parallel to  x+2y-3 = 0                    (1)

is x+2y+k = 0                                                                                 (2)

Equation (2) passes through (-1,4)

put x=-1, y =4 in equation (2)

(-1) + 2(4) +k = 0

-1 + 8 + k = 0

k = -7

∴ Required line is x +2y – 7 = 0

4.        Find the equation of the  line passing through (3,- 3)  and

perpendicular  to 4x – 3y – 10 = 0.

Soln:   Let the equation of line perpendicular to  4x – 3y – 10 = 0               (1)

is – 3x  – 4y + k = 0                                                                                       (2)

Equation (2) passes through (3,-3)

put x = 3, y =- 3  in equation (2)

-3 (3) – 4(-3) +k = 0

-9 + 12+ k = 0

k = – 3

– 3x  – 4y  – 3 = 0

∴ Required line is 3x +4y + 3 = 0

PAIR OF STRAIGHT LINES

Pair of Straight lines passing through origin

The general equation of pair of straight lines passing through origin is

ax2  +  2hxy  + by2 =  0         ———-   ( A )

Let   y = m1x

y – m1x =  0 ————–  ( 1 )

and  y = m2x

y – m2x =  0 ————–  ( 2 )

(y – m1x) (y – m2x )  =  0

y2  – (m1 +  m2 ) xy   +   m1 m2 x2  =  0

m1 m2 x2  – (m1 +  m2 ) xy   + y2  =  0  ————–  ( B )

Equations (A)  and  ( B ) represent the same pair of straight lines. Hence the ratios

of  the corresponding  coefficients of like terms are proportional.

m1 m2 / a   =   – (m1 +  m2 ) / 2h    =  1 / b

m1 +  m2  =  – 2h / b               and      m1 m2  =  a / b

i.e  sum of slopes  =  – 2h / b    and   product of slopes  =  a / b

Formulae

1 )  The angle between the pair of straight lines ax2  +  2hxy  + by2 =  0  is

tanθ   =   ± (2√h2 –  ab  ) / a + b

2)     The condition for the pair of lines to be parallel is   h2 –  ab = 0.

3)     The condition for the pair of lines to be perpendicular is   a + b = 0.

Part – A

1. Write down the combined equation of the pair of lines

x − 2y  = 0  and  3x + 2y = 0

Soln:    The two separate lines are x − 2y  = 0  and  3x + 2y = 0

The combined equation is

(x − 2y) (3x + 2y)  = 0

3x2 + 2xy – 6xy  – 4 y2 =  0

3x– 4xy  – 4 y2 =  0

2.        Write down the separate equations of the pair of lines 12x+ 7xy  – 10y2 =  0

Soln:   12x+ 7xy  – 10y2 =  0

12x+ 15xy  – 8xy – 10y2 =  0

3x( 4x + 5 )  –  2y( 4x – 5y)  = 0

(3x – 2y)  ( 4x – 5y)   =  0

∴ The separate equations are3x – 2y  =  0  and    4x – 5y  = 0

3.         Show that the two lines represented by 4x+ 4xy  + y2 =  0  are parallel to each

other.

Soln:    4x+ 4xy  + y2 =  0

This is of the form   ax2  +  2hxy  + by2 =  0

a = 4,   2h = 4     h = 2,   b  = 1

If the lines are parallel then     h2 –  ab = 0

h2 –  ab = (2)2 –  (4) ( 1 )

=  4  –  4

=   0

∴  pair of lines are parallel

4.          Find the value of  ‘p’  if the pair of  lines 4x+ pxy  + 9y2 =  0  are parallel to each

Other.

Soln:    4x+ pxy  + 9y2 =  0

This is of the form   ax2  +  2hxy  + by2 =  0

a = 4,   2h = p     h = p / 2,   b  = 9

Given lines are parallel

h2 –  ab = 0

( p /2)2  – ( 4 ) ( 9 )  =  0

P2 / 4   –  36    =  0

P2  =  144

∴  p  =  ± 12

5.    Prove that the two lines represented by 7x– 48xy  – 7y2 =  0  are perpendicular to

each  Other.

Soln:    7x– 48xy  – 7y2 =  0

This is of the form   ax2  +  2hxy  + by2 =  0

a = 7,   2h = – 48     h = – 24,   b  = – 7

If the lines are perpendicular  then     a + b = 0

a + b = 7 – 7

=   0

∴  pair of lines are perpendicular.

6.      .    Find the value of  ‘p’  if the pair of  lines px– 5xy  + 7y2 =  0  are perpendicular

to each  Other.

Soln:    px– 5xy  + 7y2 =  0

This is of the form   ax2  +  2hxy  + by2 =  0

a = p,   2h = – 5     h = – 5 / 2,   b  = 7

Given lines are perpendicular

a + b = 0

p  +  7  =  0

∴  p  =  – 7

Part – B

1.     Find the separate equations of the line 2x– 7xy  + 3y2 =  0.  Also find the angle

between them.

Soln:   2x– 7xy  + 3y2 =  0

2x– 6xy  – xy + 3y2 =  0

2x(x – 3y )  –  y( x – 3y)  = 0

(2x – y)  ( x – 3y)   =  0

∴ The separate equations are 2x – y  =  0  and    x – 3y  = 0

2x– 7xy  + 3y2 =  0

This is of the form   ax2  +  2hxy  + by2 =  0

a = 2,   2h = – 7    h = – 7 / 2,   b  = 3

Let θ be the angle between the two straight lines

W.K.T   tanθ   =   ± (2√h2 –  ab ) / a + b

=   ± (2√ (( – 7 / 2)2 –  (2) (3)  )/ (2 + 3)

=   ± (2 √ (49 / 4 –  6 ) / 5

=    ± (2 √ (49 – 24) / 4  ) / 5

=   ± (2 √ 25 / 4 ) / 5

=   ± (2 ( 5 / 2 ) / 5

tanθ   =  ± 1

tan θ = tan 45,            ∴ θ = 45°

2.    The slope of one of the lines ax2  +  2hxy  + by2 =  0  is thrice that of the other.  Show that 3h2= 4ab

Soln:   ax2  +  2hxy  + by2 =  0        ——————–  ( 1 )

Let  y = m1x   and   y = m2x  be the separate equations of equation ( 1 )

m1+  m2  =  – 2h / b              ———————  ( 2 )

m1m2  =  a / b                        ———————-  ( 3 )

Slope of one of the line = thrice slope of the other line

i.e   m1=  3m2

Equation (2) becomes

3m2 +  m2  =  – 2h / b

4m2  =  – 2h / b

m2  =  – 2h / 4b

m2  =  – h / 2b

Substitute  m1=  3m2  in equation (3 )

3m2m2  =  a / b

3 (m2 )2  =  a / b

3   (- h / 2b )2  =  a / b

3   (h2 / 4b2 )  =  a / b

3h2 / 4b2  =  a / b

3h2b    =  4ab2

(i.e)  3h2 =  4ab

PAIR OF STRAIGHT LINES NOT PASSING THROUGH THE ORIGIN

Formulae

1)  The angle between the pair of straight lines  ax2  +  2hxy  + by2 + 2gx + 2fy + c = 0

is    tanθ   =   (2√h2 –  ab ) / a + b

2)     The condition for the pair of lines to be parallel is   h2–  ab = 0.

3)     The condition for the pair of lines to be perpendicular is   a + b = 0.

4)   Condition for the second degree equation ax2  +  2hxy  + by2 + 2gx + 2fy + c = 0  to

represent  a pair of straight lines is

abc  +  2fgh – af2 – bg2 – ch2 = 0

Part – C

1.     Show that the equation  3x2  +  7xy  + 2y2 + 5x + 5y + 2= 0  represents a pair of

straight lines.

Soln:  Given  3x2  +  7xy  + 2y2 + 5x + 5y + 2= 0

Comparing with  ax2  +  2hxy  + by2 + 2gx + 2fy + c = 0

a = 3,   2h = 7           ,  b = 2,    2g = 5       ,    2f = 5                 ,   c = 2

h = 7/ 2                     g = 5/ 2           f = 5/ 2

To show  abc  +  2fgh – af2 – bg2 – ch2 = 0

abc  +  2fgh – af2 – bg2 – ch2 =

(3) (2) (2)  + 2 (5/2) (5/2) (7/2) – 3 (5/ 2)2  – 2 (5/ 2)2  – 2 (7/ 2)2

=   12  +  175/4   –  75/4  – 50/4  – 98/4

=   12  +  ( 175 – 75 – 50 – 98)/4

=   12 + ( 175 – 223)/4

=  12  +   (  -48 /4 )

=   12  – 12  =  0

Hence, the given equation represents pair of straight lines.

2.           Find K if   2x2  –  7xy  + 3y2 + 5x – 5y + k= 0  represents a pair of

straight lines.

Soln:  Given  2x2  –  7xy  + 3y2 + 5x – 5y + k= 0 ————– ( 1 )

Comparing with  ax2  +  2hxy  + by2 + 2gx + 2fy + c = 0

a = 2,   2h = – 7           ,  b = 3,    2g = 5       ,    2f = – 5                 ,   c = k

h = – 7/ 2                     g = 5/ 2           f =  – 5/ 2

Given equation ( 1 ) represents a pair of straight lines

i.e     abc  +  2fgh – af2 – bg2 – ch2 =   0

(2) (3) (k)  + 2 (- 5/2) (5/2) (-7/2) – 2 (- 5/ 2)2  – 3 (5/ 2)2  – k (- 7/ 2)2 = 0

6k  +  175/4   –  50/4  – 75/4  – 49k/4   =  0

(24k – 49k )/4  +  ( 175 – 50 – 75)/4  =  0

– 25k/4 + 50/4  =  0

( -25k  + 50 ) / 4  = 0

-25 k  = – 50

k  =  2