Question

If the slope of one of the lines given by $a{x}^2+2hxy+b{y}^2=0$ is square of the slope of the other line, then show that $a^{2}b+a{b}^2+8h^2=6abh$.

Answer 1

let $m$ be the slope of one of the lines given by $a{x}^2+2hxy+b{y}^2=0$

Then the other line has slope $m^2$

$\therefore m+m^2=\frac{-2h}{b}...\left(1\right)$ and

$\left(m\right)\left(m^2\right)=\frac{a}{b}$

i.e. $m^3=\frac{a}{b}...\left(2\right)$

$\therefore (m+m^2{)}^3=m^3+(m^2{)}^3+3\left(m\right)\left(m^2\right)(m+m^2)...[\because (p+q{)}^3=p^3+q^3+3pq(p+q)]$

$\therefore {\left(\frac{-2h}{b}\right)}^3=\frac{a}{b}+\frac{a^2}{b^2}+3\frac{a}{b}\left(\frac{-2h}{b}\right)$

$\therefore \frac{-8h^2}{b^3}=\frac{a}{b}+\frac{a^2}{b^2}-\frac{6ah}{b^2}$

Multiplying by $b^3$, we get,

$-8h^3=a{b}^2+a^{2}b-6abh$

$\therefore a^{2}b+a{b}^2+8h^3=6abh$

This is the required condition.