Question

If one of the lines denoted by the line pair $a{x}^2+2hxy+b{y}^2=0$ bisects the angle between coordinate axes, then prove that $(a-b{)}^2=4h^2$.

Answer 1

Equation of pair of lines: $a{x}^2+2hxy+b{y}^2=0$

If $a{x}^2+2hxy+b{y}^2=0$ bisects the coordinate axis then the points $(x,x)$ and $(-x,x)$ will satisfy the given equation of pair of lines.

Case I:- The point $(x,x)$ satisfy the equation-

$a{x}^2+2hx.x+b{x}^2=0$

$\Rightarrow x^2(a+2h+b)=0$

$\Rightarrow a+b=-2h.....\left(1\right)$

Case II:- The point $(x,-x)$ satisfy the equation-

$a{x}^2+2hx.(-x)+b{(-x)}^2=0$

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

$\Rightarrow a+b=+2h.....\left(2\right)$

Now from ${eq}^n\left(1\right)\&\left(2\right)$, we have

$a+b=\pm 2h$

Squaring both sides, we have

${(a+b)}^2=4h^2$

Hence proved.