Question

If y = mx be one of the bisectors of the angle between the lines $a{x}^2-2hxy+b{y}^2=0$, then

• A. $h(1-m^2)+m(a-b)=0$
• B. $h(1-m^2)+m(a+b)=0$
• C. $h(1-m^2)+m(a-b)=0$
• D. $h(1+m^2)+m(a+b)=0$

Answer 1

The correct option is C

$h(1-m^2)+m(a-b)=0$

Here equation of one bisector of angle is y - mx = 0, therefore equation of second is x + my = 0.

Hence combined equation is (x + my)(y - mx) = 0

$\Rightarrow -m{x}^2-xy(m^2-1)+m{y}^2=0$ ....................(i)

Also equation of bisectors of $a{x}^2-2hxy+b{y}^2=0$ is

$-h{x}^2-(a-b)xy+h{y}^2=0$ ................(ii)

Hence (i) and (ii) are the same equations , therefore

$\frac{m}{h}=\frac{m^2-1}{(a-b)}\Rightarrow h(m^2-1)=m(a-b)$

$\Rightarrow m(a-b)+h(1-m^2)=0$