Question

Find $\frac{1}{k}$ if one of the lines given by $k{x}^2-5xy-6y^2=0$ is $4x+3y=0.$

Answer 1

Given joint equation of the line is $k{x}^2-5xy-6y^2=0$

It is in the form of $a{x}^2+2hxy+b{y}^2=0a=k;h=-\frac{5}{2};b=-6$

The auxiliary equation of the given line is of the form $b{m}^2+2hm+a=0=>(-6)m^2-5m+k=0=>6m^2+5m-k=\mathrm{0....}\left(i\right)$

One of the line is $4x+3y=0$. Its slope is $-\frac{4}{3}$. Now, $-\frac{4}{3}$ must be one of the roots of the auxiliary equation.

Substituting $m=-\frac{4}{3}$ in equation (i)

$6{\left(\frac{-4}{3}\right)}^2+5\left(\frac{-4}{3}\right)-k=0=>6\left(\frac{16}{9}\right)+5\left(\frac{-4}{3}\right)-k=0=>32-20-3k=0=>-3k=-12=>k=4\therefore \frac{1}{k}=\frac{1}{4}$