• # question_answer If the bisectors of the angles between the pairs of lines given by the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ and $a{{x}^{2}}+2hxy+b{{y}^{2}}+\lambda ({{x}^{2}}+{{y}^{2}})=0$ be coincident, then $\lambda =$ A)            a     B)            b C)            $h$                                         D)            Any real number

Bisectors of $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ are            $\frac{{{x}^{2}}-{{y}^{2}}}{a-b}=\frac{xy}{h}$                                                        .....(i)            and of $a{{x}^{2}}+2hxy+b{{y}^{2}}+\lambda ({{x}^{2}}+{{y}^{2}})=0$            i.e., $(a+\lambda ){{x}^{2}}+2hxy+(b+\lambda ){{y}^{2}}=0$are            $\frac{{{x}^{2}}-{{y}^{2}}}{(a+\lambda )-(b+\lambda )}=\frac{xy}{h}$                                        .....(ii)            Which is the same equation as equation (i). Hence for any $\lambda$belonging to real numbers, the lines will have same bisectors.