Question

If the lines represented by $x^2-2pxy-y^2=0$ are rotated about the origin through an angle $\theta ,$ one in clockwise direction and other in anti-clockwise direction, then the equation of the bisector of the angle between the lines in the new positions is

• A. $p{x}^2+2xy-p{y}^2=0$
• B. $p{x}^2+2xy+p{y}^2=0$
• C. $x^2-2pxy-y^2=0$
• D. None of these

Answer 1

Answer: (A)

$p{x}^2+2xy-p{y}^2=0$

Given eq

$x^2-2pxy-y^2=0$

comparing above eq with general form of eq $a{x}^2+2hxy+b{y}^2=0$

$a=1,h=-p,b=-1$

Now the line is rotated one in clockwise and other is anticlockwise so the both eq are replaced by each other and form the same eq as it was so here we finding eq of angle bisector by formula

$\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

$\frac{x^2-y^2}{1+1}=\frac{xy}{-p}$

$\frac{x^2-y^2}{2}=\frac{xy}{-p}$

$(-p)(x^2-y^2)=2xy$

$-p{x}^2+p{y}^2=2xy$

$p{x}^2+2xy-p{y}^2=0$