Let the equation of a curve be $ax^2 + 2hxy + by^2 + 2gx +2fy+c ....eq (i)$
and a straight line be $lx + my +n=0....eq(ii)$
Now, joint equation of line $OP$ and $OQ$ joining the origin and points of intersection $P$ and $Q$ can be obtained by making equation $(i)$ homogeneous with the help of equation of the line. Thus required equation is given by:
$ ax^2 +2hxy+by^2 +2(gx+fy)(\frac {lx+my}{-n})+c(\frac {lx+my}{-n})^2 =0$
Solving this further gives us a homogeneous equation of pair of straight lines:
$(an^2+2gln+cl^2)x^2 +2(hn^2+gmn+fln+clm)xy +(bn^2+2fmn+cm^2)y^2=0$
And it's given that all points which satisfy $eq (i)$ and $eq(ii)$ simultaneously will satisfy this new equation.
Now I have several doubts with this above concept. First doubt is that why are we fusing the two equations like that? Is there a way to fuse the two equations? Like above, should we insert the eq of straight line in the second last term and the square of it to the constant term? Basically, some help with the methodology about fusing the two eqns will help.
My second doubt is that after the two equations of the curve and the straight line are fused, we get a new set of equation which is of a pair of straight lines passing through the origin. Can anyone tell me why? I am lacking a bit of intuition here.
My last question is that basically any point lying on the curve OR the straight line OR the pair of straight lines will satisfy the third new fused equation? Right?
By homogeneous equation we mean that the sum of powers of $x$ and $y$ is the same in all terms e.g.; $Ax^2+2Hxy+By^2=0.$ Next if we put $y=Mx$, we get a quadratic equation of $M$. if these roots are real, then we have two lines passing through origin. For instance homogeneous equation $x^2+xy+y^2=0$ does not represent a pair of straight lines as $M^2+M+1=0$ doesn't give real roots. But $y^2+xy-x^2=0$ does give real roots as $M=\frac{-1\pm\sqrt{5}}{2}$ so there are two real lines passing through origin.
The homogenisation of the given quadratic by the given line is unique and you have got it correctly. if you put $y=Mx$and your last equation gives real roots for $M$, then you get two lines passing from origin.