Question

If one of the lines given by $a{x}^2+2hxy+b{y}^2=0$ is perpendicular to $px+qy=0$ , show that $a{p}^2+2hpq+b{q}^2=0$

Answer 1

$a{x}^2+2hxy+b{y}^2=0⟶$ $\left(i\right)y-m_{1}x=0\left(ii\right)y-m_{2}x=0$
Taking conman of $x^2.$
$\frac{b{y}^2}{x^2}+\frac{2hy}{x}=0⟶$ $\left(i\right)m_1\left(ii\right)m_2$
$m_1+m_2=-\frac{2h}{b}{m}_1{m}_2=\frac{a}{b}$
Given line $px+qy=0m=\frac{-p}{q}$
Let's say $m_1\perp m{m}_1.m=-1m_1\times \frac{-p}{q}=-1m_1=\frac{q}{p}$
Putting the value of slope in eq $\left(i\right)$
$b.\frac{q^2}{p^2}+2h.\frac{q}{p}+a=0b{q}^2+2hpq+a{p}^2=0$
(Hence proved)