Question

If θ is the acute angle between the lines represented by equation ax² + 2hxy + by² = 0  then prove that `tantheta=|(2sqrt(h^2-ab))/(a+b)|, a+b!=0`

Answer 1

Case (1)

Let m₁ and m₂ are the slopes of the lines representedby
the equation ax²+2hxy+by²=0,
then  m₁ + m_{2 =}-2h/b and m₁m₂ =a/b
If θ is the acute angle between the lines,
then `tantheta=|(m_1-m_2)/(1+m_1m_2)|`

`"now " (m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2`

`(m_1-m_2)^2=((-2h)/b)^2-4(a/b)`

`(m_1-m_2)^2=(4(h^2-ab))/b^2`

`|m_1-m_2|=|(2sqrt(h^2-ab))/b|`

`"similarly "1+m_1m_2=1+a/b=(a+b)/b`

`"substituting in "tantheta=|(m_1-m_2)/(1+m_1m_2)| , " we get "`

`tantheta=|((2-sqrt(h^2-ab))/b)/((a+b)/b)|`

`tantheta=|(2-sqrt(h^2-ab))/(a+b)| , "if "a+b!=0`

 

Case (2)
If one of the lines is parallel to the y-axis then one of the slopes m₁,m₂, does not exist. As the line passes through the origin so one line parallel is the y-axis, it's equation is
x=0 and b=0
The other line is ax+2hy=0 whose slope `tab beta=-a/(2h)`

∴ The acute angle between the pair of lines is `pi/2-beta`

`therefore tantheta=|tan(pi/2-beta)|=|cot beta|=|(2h)/a|`

`"put " b=0 " in " tantheta=|(2sqrt(h^2-ab))/(a+b)|, " we get " tantheta=|(2h)/a|`

`"Hence "tantheta=|(2sqrt(h^2-ab))/(a+b)| " is valid in both the cases."`