ax2+2hxy+by2=0
Let the lines represented by the given equation be y=m1x...(i) and y=m2x....(ii)
m1+m2=−2hbm1m2=ab
Given line xcosα+ysinα=p......(iii)
A be the point of intersection (i) and (iii)
Solving the equations we get x=pcosα+m1sinα,y=m1pcosα+m1sinα
∴A(pcosα+m1sinα,m1pcosα+m1sinα)
Solving (ii) and (iii)
we get x=pcosα+m2sinα,y=m2pcosα+m2sinα
∴B(pcosα+m2sinα,m2pcosα+m2sinα)
Area of triangle △OAB
=12∣∣ ∣ ∣ ∣ ∣∣001pcosα+m1sinαm1pcosα+m1sinα1pcosα+m2sinαm2pcosα+m2sinα1∣∣ ∣ ∣ ∣ ∣∣=12p2(m2−m1)(cosα+m1sinα)(cosα+m2sinα)=p22√(m2+m1)2−4m1m2cos2α+(m2+m1)cosαsinα+m1m2sin2α=p22.√4h2b2−4abcos2α+(−2hb)cosαsinα+absin2α=p2√h2−abbcos2α−2hcosαsinα+asin2α
Also area of triangle △OAB =12(OM)(AB)
OM=|0.cosα+0.sinα−p|√cos2α+sin2α=p
12p(AB)=p2√h2−abbcos2α−2hcosαsinα+asin2α⇒AB=2p√h2−abbcos2α−2hcosαsinα+asin2α
So the distance between the point of intersection is 2p√h2−abbcos2α−2hcosαsinα+asin2α
Hence proved.