Question

Show that the pair
$a{x}^2+2hxy+b{y}^2+\lambda (x^2+y^2)=0$
is equally inclined to the same pair.

Answer 1

The equation of angle bisector of homogeneous equation of second degree is $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

$a{x}^2+2hxy+b{y}^2+\lambda (x^2+y^2)=0(a+\lambda )x^2+2hxy+(b+\lambda )y^2=0$

Equation of angle bisector

$\frac{x^2-y^2}{(a+\lambda )-(b+\lambda )}=\frac{xy}{h}\Rightarrow \frac{x^2-y^2}{a-b}=\frac{xy}{h}$

This gives the same equation for the angle bisector irrespective the value of $\lambda$

Thus it will be equally inclined with the same pair .