# 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?

Aug 8, 2018

The given eqyation of a pair of straight line is $a {x}^{2} + 2 h x y + b {y}^{2} + \lambda \left({x}^{2} + {y}^{2}\right) = 0$

$\mathmr{and} , \left(a + l a m \mathrm{da}\right) {x}^{2} + 2 h x y + \left(b + \lambda\right) {y}^{2} = 0$

$\mathmr{and} , c {x}^{2} + 2 h x y + {\mathrm{dy}}^{2} = 0$ ,

Where $a + \lambda = c \mathmr{and} b + \lambda = d$

The equation of the pair of bisectors becomes

$h \left({x}^{2} - {y}^{2}\right) = \left(c - d\right) x y$

$\mathmr{and} , h \left({x}^{2} - {y}^{2}\right) = \left(a + \lambda - b - \lambda\right) x y$

$\mathmr{and} , h \left({x}^{2} - {y}^{2}\right) = \left(a - b\right) x y$

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