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#?

Question

7036 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-08T15:35:51

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#