# To Find Constant Using Homogeneous Equation

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### Type – IV B: To Find Value of Constant Using Auxiliary Equation of the Pair of lines

#### ALGORITHM:

• Write the auxiliary equation of joint equation.
• Find the slope of given line m.
• Substitute value of m in auxillary equation.
• Simplify and get the value of constant.

#### Example – 12:

• Find the value of ‘k’ if one of the line given by 6x2 + kxy + y2 = 0 is 2x + y = 0.
• Solution:

Given joint equation of lines is 6x2 + kxy + y2 = 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = 6, 2h = k and b= 1

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴   (1)m2 + km + 6 = 0

∴   m2 + km + 6 = 0 ……… (1)

One of the line is 2x + y = 0. Its slope is – 2/1 = -2

Now – 2 must be one of the roots of the auxiliary equation (1),

Substituting m = – 2 in equation (1)

(-2)2 + k(-2) + 6 = 0

∴   4 – 2k + 6 = 0

∴   – 2k  = – 10

∴   k = 5

#### Example – 13:

• Find λ, if  2x + 5y = 0 coincides with one of the lines x2 – λxy +  5y2 = 0.
• Solution:

Given joint equation of lines is x2 – λxy +  5y2 = 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = 1, 2h = – λ and b= 5

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴   (5)m2 – λm + 1 = 0

∴   5m2 – λm + 1 = 0 ……… (1)

One of the line is 2x + 5y = 0. Its slope is – 2/5

Now – 2/5 must be one of the roots of the auxiliary equation (1),

Substituting m = – 2/5 in equation (1)

5(-2/5)2 –  λ(-2/5) + 1 = 0

∴  5(4/25) –  λ(-2/5) + 1 = 0

∴  (4/5) –  λ(-2/5) + 1 = 0

Multiplying both sides of equation by 5

∴  4 +  2λ  + 5 = 0

∴   2λ = – 9

∴  λ = – 9/2

#### Example – 14:

• Find k, if the one of the lines given by 2x2 – xy  + ky2 = 0 is x – 3y = 0 .
• Solution:

Given joint equation of lines is 2x2 – xy  + ky2 = 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = 2, 2h = – 1 and b= k

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴   (k)m2 – 1m + 2 = 0

∴   km2 –  m + 2 = 0 ……… (1)

One of the line is x – 3y = 0. Its slope is – 1/-3 = 1/3

Now 1/3 must be one of the roots of the auxiliary equation (1),

Substituting m = 1/3 in equation (1)

k(1/3)2 –  (1/3) + 2 = 0

∴  K(1/9) –  1/3 + 2 = 0

Multiplying both sides of equation by 9

∴  k – 3 + 18 = 0

∴   k = -15

#### Example – 15:

• Find k, if the one of the lines given by kx2 + 3xy  – y2 = 0 is 2x + y = 0
• Solution:

Given joint equation of lines is kx2 + 3xy  – y2 = 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = k, 2h = 3 and b = – 1

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴   (-1)m2 + 3m + k = 0

∴   m2 –  3m – k = 0 ……… (1)

One of the line is 2x + y = 0. Its slope is – 2/1 = -2

Now -2 must be one of the roots of the auxiliary equation (1),

Substituting m = -2 in equation (1)

(-2)2 –  3(-2) – k = 0

∴  4 + 6 – k = 0

∴  k = 10

#### Example – 16:

• Find k, if the one of the lines given by 6x2 – 14xy  + 14ky2 = 0 coincides with y = 2x
• Solution:

Given joint equation of lines is 6x2 – 14xy  + 14ky2 = 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = 6, 2h = – 14 and b = 14k

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴   (14k)m2 – 14 m + 6 = 0

∴  14km2 – 14 m + 6 = 0 ……… (1)

One of the line is y = 2x. Its slope is 2

Now 2 must be one of the roots of the auxiliary equation (1),

Substituting m = 2 in equation (1)

14k(2)2 – 14 (2) + 6 = 0

∴  56 k – 28 + 6 = 0

∴  56 k  = 22

∴   k  = 22/56 = 11/28

#### Example – 17:

• Find k, if the one of the lines given by kx2 – 5xy  – 6y2 = 0 is 4x + 3y = 0
• Solution:

Given joint equation of lines is kx2 – 5xy  – 6y2 = 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = k, 2h = – 5 and b = – 6

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴   (- 6)m2 – 5m + k = 0

∴  6m2 + 5 m – k = 0……… (1)

One of the line is 4x + 3y = 0. Its slope is – 4/3

Now – 4/3  must be one of the roots of the auxiliary equation (1),

Substituting m = – 4/3 in equation (1)

6(- 4/3)2 + 5 (- 4/3) – k = 0

∴  6(16/9) + 5 (- 4/3) – k = 0

Multiplying both sides of equation by 3

∴  32 – 20 – 3k = 0

– 3k = – 12

∴   k  = 4

#### Example – 18:

• Find k, if the one of the lines given b 3x2 + kxy  + 2y2 = 0 is 2x + y = 0
• Solution:

Given joint equation of lines is 3x2 + kxy  + 2y2 = 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = 3, 2h = k and b = 2

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴  2m2 + k m + 3 = 0 ……… (1)

One of the line is 2x + y = 0. Its slope is – 2/1 = -2

Now – 2  must be one of the roots of the auxiliary equation (1),

Substituting m = – 2 in equation (1)

2(-2)2 + k (-2) + 3 = 0

∴  8 – 2k + 3 = 0

∴  – 2k = – 11

∴   k  = 11/2

Example – 19:

• Find the value of ‘k’ if one of the line given by 4x2 + kxy – y2 = 0 is 2x + y = 0.
• Solution:

Given joint equation of lines is 4x2 + kxy – y2 = 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = 4, 2h = k and b = -1

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴  (-1)m2 + k m + 4 = 0

∴  m2 – k m – 4 = 0  ……… (1)

One of the line is 2x + y = 0. Its slope is – 2/1 = -2

Now – 2  must be one of the roots of the auxiliary equation (1),

Substituting m = – 2 in equation (1)

(-2)2 – k (-2) – 4 = 0

∴  4 + 2k  – 4 = 0

∴  2k = 0

∴   k  = 0

#### Example – 20:

• Find the value of ‘a’ if the line given by ax2 + xy – 3y2 = 0 is perpendicular to 3x – 5y – 1 = 0.
• Solution:

Given joint equation of lines is ax2 + xy – 3y2 = 0

It is in the form Ax2 + 2Hxy  + By2 = 0.

A = a, 2H = 1 and B = – 3

The auxiliary equation of given line is of the form

Bm2 + 2Hm + A = 0

∴  (-3)m2 + 1 m + a = 0

∴  3m2 – m – a = 0  ……… (1)

Given line is 3x – 5y – 1 = 0. slope of this line is – 3/-5 = 3/5

One of the line is perpendicular to 3x – 5y – 1 = 0. Hence its slope = – 5/3

Now – 5/3  must be one of the roots of the auxiliary equation (1),

Substituting m = – 5/3 in equation (1)

∴  3(- 5/3)2 – (- 5/3) – a = 0

∴  3(25/9) + (5/3) – a = 0

Multiplying both sides of equation by 3

∴  25 + 5 – 3a = 0

∴   – 3a = – 30

∴   a  = 10

#### Example – 21:

• Find the value of ‘k’ if the line given by 2x2 – 5xy + ky2= 0 is perpendicular to x – 2y = 8.

Solution:

Given joint equation of lines is 2x2 – 5xy + ky2= 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = 2, 2h = – 5 and b = k

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴ km2 – 5m + 2 = 0  ……… (1)

Given line is x – 2y = 8. slope of this line is – 1/-2 = 1/2

One of the line is perpendicular to x – 2y = 8. Hence its slope = – 2

Now – 2  must be one of the roots of the auxiliary equation (1),

Substituting m = – 2 in equation (1)

∴  k(-2)2 – 5(-2) + 2 = 0

∴  4 k + 10 + 2 = 0

∴   4k = – 12

∴   k  = – 3

#### Example – 22:

• Find the value of ‘k’ if the line given b 3x2 – kxy + 5y2= 0 is perpendicular to 5x + 3y = 0.

Solution:

Given joint equation of lines is 3x2 – kxy + 5y2= 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = 3, 2h = – k and b = 5

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴  5m2 – km + 3 = 0  ……… (1)

Given line is 5x + 3y = 0. slope of this line is – 5/3

One of the line is perpendicular to 5x + 3y = 0. Hence its slope =  3/5

Now  3/5  must be one of the roots of the auxiliary equation (1),

Substituting m =  3/5 in equation (1)

∴ 5(3/5)2 – k(3/5) + 3 = 0

∴ 5(9/25) – k(3/5) + 3 = 0

Multiplying both sides of equation by 5

∴  9 – 3k + 15 = 0

∴   – 3k = – 24

∴   k  = 8

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