Question

If the slope of one of the lines represented by $a{x}^2+2hxy+b{y}^2=0$ be the square of the other,
then prove that $\frac{a+b}{h}+\frac{8h^2}{ab}=6$.

Answer 1

If the slop of the two lines be $m,m^2$ then
$m+m^2=-\frac{2h}{b}$ and $m,m^2=\frac{a}{b}$
Cubing first relation, we have
$m^3+m^6+3m.m^2(m+m^2)=-\frac{8h^3}{b^3}$
$\therefore \frac{a}{b}+\frac{a^2}{b^2}+\frac{3a}{b}(-\frac{2h}{b})=-\frac{8h^3}{b^3}$
or $\frac{a(a+b)}{b^2}-\frac{6ah}{b^2}=-\frac{8h^3}{b^3}$
Multiply throughout by $\frac{b^2}{ah}$
$\frac{a+b}{h}+\frac{8h^2}{ab}=6$.