Question

The line $x-2y=0$ will be a bisector of the angle between the lines represented by the equation $x^2-2hxy-2y^2=0$, if $h=$

• A. 1/2
• B. 2
• C. -2
• D. -1/2

Answer 1

The correct option is A ($2$)

Here, one equation of bisectors is $x-2y=0$ compare with $ax+by=0$, we get $a=1,b=-2$.
Bisector of homogenous pair of line is $h{x}^2-(a-b)xy-h{y}^2=0$
Substitute $a=1,b=-2$ in above equation, we get
$\Rightarrow h{x}^2-(1+2)xy-h{y}^2=0$
$\Rightarrow h{x}^2-3xy-h{y}^2=0$
$\Rightarrow h(x^2-y^2)-3xy=0$
$\Rightarrow h(x^2-y^2)=3xy$
$\Rightarrow \frac{(x^2-y^2)}{3}=\frac{xy}{h}$
Substitute given equation $y=2x$ in above equation, we get
$\Rightarrow \frac{4y^2-y^2}{3}=\frac{2y^2}{h}$
$\Rightarrow \frac{3y^2}{3}=\frac{2y^2}{h}$
$\Rightarrow h=\frac{2y^2}{y^2}$
$\therefore h=2$