Question

Prove that the straight lines $a{x}^2+2hxy+b{y}^2=0$ have the same pair of bisectors as those of the line given by $a^2{x}^2+2h(a+b)xy+b^2{y}^2=0$.

Answer 1

The bisectors of the pair of lines $a^2{x}^2+2h(a+b)xy+b^2{y}^2=0$ is given by

$\frac{x^2-y^2}{coeff.of{x}^2-coeff.of{y}^2}=\frac{xy}{halfthecoeff.ofxy}$

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

But these are also the bisectors of the lines given by first pair.