Question

Show that the straight lines joining the origin to the other two points of intersection of the curves whose equations are
$a{x}^2+2hxy+b{y}^2+2gx=0$
and $a^{\prime }{x}^2+2h^{\prime }xy+b^{\prime }{y}^2+2g^{\prime }x=0$
will be at right angles if $g(a^{\prime }+b^{\prime })-g^{\prime }(a+b)=0.$

Answer 1

$a{x}^2+2hxy+b{y}^2+2gx=\mathrm{0.....}\left(i\right)a^{\prime }{x}^2+2h^{\prime }xy+b^{\prime }{y}^2+2g^{\prime }x=\mathrm{0.......}\left(ii\right)$

$2gx=-(a{x}^2+2hxy+b{y}^2)2x=\frac{-(a{x}^2+2hxy+b{y}^2)}{g}$

Substituting $2x$ in $\left(ii\right)$
$a^{\prime }{x}^2+2h^{\prime }xy+b^{\prime }{y}^2+g^{\prime }\left(\frac{-(a{x}^2+2hxy+b{y}^2)}{g}\right)=0g{a}^{\prime }{x}^2+2g{h}^{\prime }xy+g{b}^{\prime }{y}^2-g^{\prime }a{x}^2-2g^{\prime }hxy-g^{\prime }b{y}^2=0(g{a}^{\prime }-g^{\prime }a)x^2+(g{b}^{\prime }-g^{\prime }b)y^2+(2g{h}^{\prime }-2g^{\prime }h)xy=0$

Lines are perpendicular if $a+b=0$

$\Rightarrow g{a}^{\prime }-g^{\prime }a+g{b}^{\prime }-g^{\prime }b=0g(a^{\prime }+b^{\prime })-g^{\prime }(a+b)=0$

Hence proved.