Question

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`?

Answer 1

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`