The discussion focuses on proving that one of the angle bisectors of the lines defined by the equation ax² + 2hxy + by² = 0 passes through the intersection point of the lines represented by ax² + 2hxy + by² + 2gx + 2fy + c = 0. The condition for this to hold is h(g² - f²) = fg(a - b). The intersection point is determined as (hf - bg)/(ab - h²), (hg - af)/(ab - h²), and the angle bisector equation is hx² + (b - a)xy - hy² = 0. By substituting the intersection point into the bisector equation, the proof is established through factorization.
PREREQUISITESStudents and educators in mathematics, particularly those studying geometry and conic sections, as well as anyone involved in algebraic proofs and geometric constructions.
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.suvadip said: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 $h(g^2-f^2)=fg(a-b).$