Question

If the equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represent a pair of straight lines, then prove that the equation to the third pair of straight lines passing through the points where these meet the axis is $a{x}^2-2hxy+b{y}^2+2gx+2fy+c+\frac{4fg}{c}xy=0$

Answer 1

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

using the concept of family of lines , the other two lines will be

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c+\lambda xy=\mathrm{0.......}\left(i\right)$

For this to represent straight lines

$\Delta =0abc+2fgh-a{f}^2-{bg}^2-{ch}^2=0ab+2fg\frac{2h+\lambda }{2}-a{f}^2-{bg}^2-c{\left(\frac{2h+\lambda }{2}\right)}^2=0(abc+2fgh-a{f}^2-{bg}^2-{ch}^2)+\lambda fg-\frac{c{\lambda }^2}{4}-ch\lambda =00+\lambda fg-\frac{c{\lambda }^2}{4}-ch\lambda =0\frac{c{\lambda }^2}{4}-\lambda fg+ch\lambda =0\lambda (\frac{c\lambda }{4}+fg-ch)=0\lambda =0,\frac{4(fg-ch)}{c}\lambda =\frac{4fg}{c}-4h$

Substituting in $\left(i\right)$

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c+(\frac{4fg}{c}-4h)xy=0\Rightarrow a{x}^2-2hxy+b{y}^2+2gx+2fy+c+\frac{4fg}{c}xy=0$

Hence proved.