Prove that the straight lines ax^2+2hxy+by^2+lambda(x^2+y^2)=0 have the same pair of bisectors for every value of lambda?

1 Answer
Aug 8, 2018

The given eqyation of a pair of straight line is ax^2+2hxy+by^2+lambda(x^2+y^2)=0

or,(a+lamda)x^2+2hxy+(b+lambda)y^2=0

or,cx^2+2hxy+dy^2=0 ,

Where a+lambda=candb+lambda=d

The equation of the pair of bisectors becomes

h(x^2-y^2)=(c-d)xy

or,h(x^2-y^2)=(a+lambda-b-lambda)xy

or,h(x^2-y^2)=(a-b)xy

This eqution is independent of lambda.
So the given pair of lines have the same pair of bisectors for every value of lambda