Question

If the slope of one of the line given by $3x^2+4xy+k{y}^2=0$ is three times the other.

Answer 1

$3x^2+4xy+k{y}^2=0$
$\therefore$ divide by $x^2$
$=\frac{3x^2}{x^2}+\frac{4xy}{x^2}+\frac{k{y}^2}{x^2}=0$
$3+\frac{4y}{x}+\frac{k{y}^2}{x^2}=\mathrm{0...}\left(1\right)$
$\therefore y=mx$
$\therefore \frac{y}{x}=m$
Put $\frac{y}{x}=m$ in equation $\left(1\right)$
Comparing the equation $k{m}^2+4m+3=0$ with $a{x}^2+2hxy+b{y}^2=0$,
we get,
$a=k,2h=4=3$
$m_1=3m_2$...(given condition)
$m_1+m_2=\frac{-2h}{k}=-\frac{4}{k}$
$m_1{m}_2=\frac{a}{b}=\frac{3}{k}$
$m_1+m_2=-\frac{4}{k}$
$4m_2=-\frac{4}{k}...(m_1=3m_2)$
$m_2=-\frac{1}{k}$
$m_1{m}_2=\frac{3}{k}$
$3m_2^2=\frac{3}{k}...(m_1=3m_2)$
$3{(-\frac{1}{k})}^2=\frac{3}{k}...(m_2=-\frac{1}{k})$
$\frac{1}{k^2}=\frac{1}{k}$
$k^2=k$
$k=1$ or $k=0$