Question: Suppose we have a curve $$\mathcal{S} :\ ax^2+by^2+2hxy+2gx+2fy+c=0$$
and a straight line $$\mathcal{L} :\ lx+my+n=0$$
where $n \ne 0$ and we have the homogenized curve of them $$\mathcal{S}' :\ ax^2+by^2+2hxy+2(gx+fy)\left(\frac{lx+my}{-n} \right)+c\left(\frac{lx+my}{-n}\right)^2=0$$
then prove or disprove the following statement : "If $\mathcal{L}$ intersects $\mathcal{S}$ at exactly one real point then $\mathcal{S}'$ represents a pair of real coincident lines passing through origin and through the point of intersection of $\mathcal{S}$ and $\mathcal{L}$."
What this statement basically concludes is that by the very definition of homogenization all the lines of $\mathcal{S'}$ must represent the line passing through origin and point of intersection of $\mathcal{S}$ and $\mathcal{L}$ but that clearly fails as in the example taken below.
My Attempt:
Now I could not formally prove anything in this but I was able to find a counterexample of this which does disprove this but I don't know the logic behind this so any explanation would be helpful, here's the example :
Let $$\mathcal{S} :\ x – 4y^2 +5 =0$$ $$\mathcal{L}:\ y=7 $$ then $$\mathcal{S}' :\ x\left(\frac{y}{7}\right)-4y^2 +5\left(\frac{y}{7}\right)^2 =0 \implies y\left(\frac{x}{7}-\frac{191}{49}y \right)=0$$
Clearly $\mathcal{S}'$ does not represent a pair of coincident lines.
Why is the line $y=0$ coming?
Best Answer
The projective situation is $\langle xz-4y^2-5z^2,y-7z\rangle$ which is $\langle x-191z,y-7z\rangle$ and $\langle y,z\rangle.$ This means that the line intersects the parabola also at the line at infinity along the $x$-axis or $y=0.$