Question

If θ is the acute angle between the lines represented by equation ax² + 2hxy + by² = 0  then prove that ${\displaystyle \tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|,a+b\ne 0}$

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 ${\displaystyle \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|}$

${\displaystyle \text{now}{(m_1-m_2)}^2={(m_1+m_2)}^2-4m_1{m}_2}$

${\displaystyle {(m_1-m_2)}^2={\left(\frac{-2h}{b}\right)}^2-4\left(\frac{a}{b}\right)}$

${\displaystyle {(m_1-m_2)}^2=\frac{4(h^2-ab)}{b^2}}$

${\displaystyle |m_1-m_2|=\left|\frac{2\sqrt{h^2-ab}}{b}\right|}$

${\displaystyle \text{similarly}1+m_1{m}_2=1+\frac{a}{b}=\frac{a+b}{b}}$

${\displaystyle \text{substituting in}\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|,\text{we get}}$

${\displaystyle \tan \theta =\left|\frac{\frac{2-\sqrt{h^2-ab}}{b}}{\frac{a+b}{b}}\right|}$

${\displaystyle \tan \theta =\left|\frac{2-\sqrt{h^2-ab}}{a+b}\right|,\text{if}a+b\ne 0}$

 

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 ${\displaystyle tab\beta =-\frac{a}{2h}}$

∴ The acute angle between the pair of lines is ${\displaystyle \frac{\pi }{2}-\beta }$

${\displaystyle \therefore \tan \theta =\left|\tan (\frac{\pi }{2}-\beta )\right|=\left|\cot \beta \right|=\left|\frac{2h}{a}\right|}$

${\displaystyle \text{put}b=0\text{in}\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|,\text{we get}\tan \theta =\left|\frac{2h}{a}\right|}$

${\displaystyle \text{Hence}\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|\text{is valid in both the cases.}}$