MHB Homework: Bisectors and Intersection Points of Straight Lines

[Suvadip]

Show that one of the bisectors of the angles between the pair of straight lines ax²+2hxy+by²=0 will pass through the point of intersection of the straight lines ax²+2hxy+by²+2gx+2fy+c=0 if
h(g²-f²)=fg(a-b)

Please help

 

[Opalg]

Show that one of the bisectors of the angles between the pair of straight lines $ax^2+2hxy+by^2=0$ will pass through the point of intersection of the straight lines $ax^2+2hxy+by^2+2gx+2fy+c=0$ if $h(g^2-f^2)=fg(a-b).$

If the conic $ax^2+2hxy+by^2+2gx+2fy+c=0$ consists of two straight lines, then the point of intersection of those lines is the centre of the conic, which is given here to be the point $$\left(\frac{hf-bg}{ab-h^2},\frac{hg-af}{ab-h^2}\right).\qquad(1)$$ The equation of the pair of angle bisectors of the lines $ax^2+2hxy+by^2=0$ is given here to be $$hx^2 + (b-a)xy -hy^2 = 0.\qquad(2)$$ Substitute the point (1) into the equation (2), and you get $$h(hf-bg)^2 + (b-a)(hf-bg)(hg-af) -h(hg-af)^2 = 0.$$ Multiply that out, and you will find that some of the terms combine or cancel. The remaining terms can be factorised to give you $$(ab-h^2)(g^2h-f^2h + bfg - afg)=0.$$ But $ab-h^2\ne0$ because that expression is always negative for a conic consisting of two straight lines (see http://www.mathhelpboards.com/f11/pair-straight-lines-3646/). Therefore the other factor must be zero, which gives the required solution.