Question

One bisector of the angle between the lines given by $a(x-1{)}^2+2h(x-1)y+b{y}^2=0$ is $2x+y-2=0$, then

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

We have,
$a(x-1{)}^2+2h(x-1)y+b{y}^2=0$
Shifting the origin at (1,0), we get
$a{x}^2+2hxy+b{y}^2=0$
Equation of bisectors
$\frac{x^2-y^2}{a-b}=\frac{xy}{h}$ .....(i)
Again, shifting the original at initial position, we get
$\frac{(x-1{)}^2-y^2}{a-b}=\frac{(x-1)y}{h}$
$(x-1{)}^2-\left(\frac{a-b}{h}\right)(x-1)y-y^2=0$ .....(ii)
One given bisector is
$2x+y-2=0$
$\Rightarrow$ Other bisector is
$x-2y+\lambda =0$
$\therefore (2x+y-2)(x-2y+\lambda )=0$
$\Rightarrow 2x^2-3xy-2y^2+(2\lambda -2)x+(\lambda +4)y-2\lambda =0$ .....(iii)
Rewriting equation (ii), as
$x^2-\left(\frac{a-b}{h}\right)xy-y^2-2x+\left(\frac{a-b}{h}\right)y+1=0$
or $2x^2-2\left(\frac{a-b}{h}\right)xy-2y^2-4x+2\left(\frac{a-b}{h}\right)y+2=0$ ....(iv)
comparing equations (iii) and (iv),we get
$2\left(\frac{a-b}{h}\right)=3,2\lambda -2=-4,2\left(\frac{a-b}{h}\right)=\lambda +4$
$2\lambda =-2$
$\therefore \frac{a-b}{h}=\frac{3}{2},\lambda =-1$