- Thread starter
- #1

- Thread starter suvadip
- Start date

- Thread starter
- #1

- Moderator
- #2

- Feb 7, 2012

- 2,785

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 this other thread). Therefore the other factor must be zero, which gives the required solution.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 $h(g^2-f^2)=fg(a-b).$