Show that the line y=mx bisects the angle between lines ax^2-2hxy+by^2=0 if h(1-m^2)+m(a-b)=0?

1 Answer
Aug 6, 2018

Let the equations y-m_1x=0andy-m_2x=0 are two straight lines represented by the given equation of pair of straight lines.Here m_1=tanalphaandm_2=tanbetaand beta>alpha

Hence

(y-m_1)(y-m_2x)=y^2-(2h)/bxy+a/bx^2

So

m_1+m_2==(2h)/bandm_1m_2=a/b

If theta be the angle subtended by angle bisector (y=mx) of the pair of straight line with the positive direction of X-axis ,then m=tantheta

Now it is obvious that

theta-alpha=beta-theta

So

alpha+beta=2theta

=>tan(alpha+beta)=tan(2theta)

=>(tanalpha+tanbeta)/(1-tanalphatanbeta)=(2tantheta)/(1-tan^2theta)

=>(m_1+m_2)/(1-m_1m_2)=(2m)/(1-m^2)

=>((2h)/b)/(1-a/b)=(2m)/(1-m^2)

=>h/(b-a)=m/(1-m^2)

=>h(1-m^2)=(b-a)m

=>h(1-m^2)+(a-b)m=0