Question

If $(a+b{)}^2=4h^2$, prove that one of the lines given by the equation $a{x}^2+2hxy+b{y}^2=0$ will bisect the angle between the co-ordinate axes.

Answer 1

Solution:-

The equation of the bisectors of the angles between the coordinate axes are $y=x$ and $y=-x$.

Let $y=x$ belong to $a{x}^2+2hxy+b{y}^2=0$

$\therefore a{x}^2+2h{x}^2+b{x}^2=0$

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

If $y=-x$ belongs to $a{x}^2+2hxy+b{y}^2=0$.

$\therefore a{x}^2-2h{x}^2+b{x}^2=0$

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

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

$(a+b)=\pm 2h$

Squaring both sides, we get

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

Hence proved.