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