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Exercise 4.5 (Solutions)page 28
Calculus and Analytic Geometry, MATHEMATICS 12 Available online @ http://www.mathcity.org, Version: 3.
Homogenous 2
nd Degree Equation
Every homogenous second degree equation 2 2 ax + 2 hxy + by = 0
represents straight lines through the origin.
Consider the equations are y = m x 1 and y = m x 2
⇒ m x 1 − y = 0 and m x 2 − y = 0
Taking product
( m x 1^ −^ y^ )( m x 2^ −^ y ) =^0
2 2 ⇒ m m x 1 2 − m xy 1 − m xy 2 + y = 0
2 2 ⇒ m m x 1 2 − m 1 (^) + m 2 (^) xy + y = 0 ………. (i)
Also we have 2 2 ax + 2 hxy + by = 0
a h x xy y b b
⇒ + + = ÷ ing by b
a h x xy y b b
Comparing it with (i), we have
1 2
a m m b
= and (^1 )
2 h m m b
Let θ be the angles between the lines then
1 2
1 2
tan 1
m m
m m
2 1 2
(^11 )
m m
m m
2 2 1 2 1 2
1 1
m m m m
m m
2 2 1 2 1 2 1 2
1 1
m m m m m m
m m
2 1 2 1 2
1 1
m m m m
m m
h a
b b
a
b
−^ −
2
2
h a
b^ b
a
b
2
2
4 h 4 ab
b
b a
b
2 4 tan
h ab
b a
2 2 tan
h ab
a b
Find the lines represented by each of the following and also find measure of the angle
between them (problem 1-6)
Question # 1
2 2 10 x − 23 xy − 5 y = 0
Solution
2 2 10 x − 23 xy − 5 y = 0 ………… (i)
2 2 ⇒ 10 x − 25 xy + 2 xy − 5 y = 0
⇒ 5 x (^) ( 2 x − 5 y (^) ) + y (^) ( 2 x − 5 y ) = 0 ⇒ (^) ( 2 x − 5 y (^) )( 5 x + y ) = 0
⇒ 2 x − 5 y = 0 and 5 x + y = 0
are the required lines.
Comparing eq. (i) with 2 2 ax + 2 hxy + by = 0
So a = 10 , 2 h = − 23
⇒ h = − , b = − 5
Let θ be angle between lines then
2 2 tan
h ab
a b
2 23 2 (10)( 5) 2
−^ −^ −
( )
tan 5
− ^
Hence acute angle between the lines = 79 31′
Question # 2
2 2 3 x + 7 xy + 2 y = 0
Solution Do yourself as above
Question # 3
2 2 9 x + 24 xy + 16 y = 0
Solution Do yourself as above
Question # 4
2 2 2 x + 3 xy − 5 y = 0
Solution
2 2 2 x + 3 xy − 5 y = 0 ………… (i)
2 2 ⇒ 2 x + 5 xy − 2 xy − 5 y = 0
⇒ x (^) ( 2 x + 5 y (^) ) − y (^) ( 2 x + 5 y ) = 0
⇒ (^) ( 2 x + 5 y (^) )( x − y ) = 0
⇒ 2 x + 5 y = 0 and x − y = 0
are the required lines.
Comparing eq. (i) with
2 2sec 4sec 4
2(1)
( )
2 2sec 4 sec 1
2 2sec 4 tan
2
2 2
∵ 1 + tan α =sec α
2sec 2 tan
2
sec tan
x
y
1 sin
cos cos
1 sin
cos
1 sin
cos
x
y
⇒ = and
1 sin
cos
x
y
⇒ x cos α = (^) ( − + 1 sinα) y and x cos α = (^) ( − − 1 sinα) y
⇒ x cos α− − + ( 1 sin α) y = 0 and x cos α− − −( 1 sin α) y = 0
⇒ x cos α+ (^) ( 1 − sin α) y = 0 and x cos α+ (^) ( 1 + sin α) y = 0
are required equations of lines.
Now comparing (i) with 2 2 ax + 2 hxy + by = 0
a = 1 , 2 h = 2sec α ⇒ h = sec α , b = 1 b
If θ is angle between lines then
2 2 tan
h ab
a b
2 2 sec (1)(1)
1 1
2 2 sec 1
2
2
= tan α
⇒ tan θ = tanα ⇒ θ = α
[
Question # 7
Find a joint equation of the lines through the origin and perpendicular to the lines:
2 2
x − 2 xy tan α− y = 0
Solution Given:
2 2
x − 2 xy tan α− y = 0
Suppose m 1 and m 2 are slopes of given lines then
1 2
2 h m m b
2 tan
1
⇒ m 1 + m 2 = −2 tan α
a m m b
⇒ m m 1 2 = − 1
Now slopes of lines ⊥ ar to given lines are 1
m
− and 2
m
− , then their equations are
a = 1 ,
2 h = −2 tan α
⇒ h = − tan α
b = − 1
1
y x m
2
y x m
= − (Passing through origin)
⇒ m y 1 = − x & m y 2 = − x
⇒ x + m y 1 = 0 & x + m y 2 = 0
Their joint equation:
( x^ +^ m y 1^ )( x^ +^ m y 2 ) =^0
( )
2 2 ⇒ x + m 1 (^) + m 2 (^) xy + m m y 1 2 = 0
( )
2 2
⇒ x + − 2 tan α xy + −( 1) y = 0
2 2
⇒ x − 2 xy tan α− y = 0
Question # 8
Find a joint equation of the lines through the origin and perpendicular to the lines:
2 2 ax + 2 hxy + by = 0
Solution
Given:
2 2 ax + 2 hxy + by = 0
Suppose m 1 and m 2 are slopes of given lines then
1 2
2 h m m b
a m m b
Now slopes of lines ⊥ ar to given lines are 1
m
− and 2
m
− , then their equations
are
1
y x m
2
y x m
= − (Passing through origin)
⇒ m y 1 = − x & m y 2 = − x
⇒ x + m y 1 = 0 & x + m y 2 = 0
Their joint equation:
( x^ +^ m y 1^ )( x^ +^ m y 2 ) =^0
( )
2 2 ⇒ x + m 1 (^) + m 2 (^) xy + m m y 1 2 = 0
h a x xy y b b
2 2 ⇒ bx − 2 hxy + ay = 0
Question # 9
Find the area of the region bounded by:
2 2 10 x − xy − 21 y = 0 and x + y + 1 = 0
Solution
2 2 10 x − xy − 21 y = 0 , x + y + 1 = 0
2 2 ⇒ 10 x − 15 xy + 14 xy − 21 y = 0
⇒ 5 x (^) ( 2 x − 3 y (^) ) + 7 y (^) ( 2 x − 3 y ) = 0
⇒ (^) ( 2 x − 3 y (^) )( 5 x + 7 y ) = 0
⇒ 2 x − 3 y = 0 or 5 x + 7 y = 0
So we have equation of lines
