Question

Prove that the pair of lines $a^2{x}^2+2h(a+b)xy+b^2{y}^2=0$ is equally inclined to the pair $a{x}^2+2hxy+b{y}^2=0$.

Answer 1

The pair of lines $a^2{x}^2+2h(a+b)xy+b^2{y}^2=0$ ....(1)

will be equally inclined to pair of lines $a{x}^2+2hxy+b{y}^2=0$ .....(2)

if the bisectors of angles between $\left(1\right)$ are same as the bisectors of the angles between $\left(2\right)$

The equation of the bisectors of the angle between line pair $\left(1\right)$ is

$h(a+b)(x^2-y^2)=(a^2-b^2)\left(xy\right)$

$\Rightarrow \frac{x^2-y^2}{a^2-b^2}=\frac{xy}{h(a+b)}$ or $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

The equation of bisector for $\left(2\right)$ is clearly $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$ which is same as $\left(1\right)$ i.e., bisector of $\left(1\right)$

Hence, the result is proved.