I need to find the equation of a parabola which touches the conic $x^2+y^2+xy-2x-2y+1=0$ at the points where it is intersected by the line $x+y+1=0$
My book says that since the parabola $(P)$ touches the conic $S=0$ $(S \equiv x^2+y^2+xy-2x-2y+1)$ at the points where it is intersected by the line $L= 0$ $(L \equiv x+y+1)$, then the parabola is in double contact with the conic at these points. So they took the equation of the parabola as
$\phi \equiv S+\lambda L^2=0$
Where does this come from? How is this true? Does it come from the fact that there are two points of intersection of the conic and the line???
Also, I graphed out the conic and the line, and from there I saw that the conic doesn't really touch the ellipse. Is the question wrong or am I just missing something here?
Best Answer
The ellipse $S=x^2+y^2+xy-2x-2y+1=0$ doesn't meet the line $x+y+1=0,$ but it does meet $L=x+y-1=0.$ Now the pencil of conics $S+\lambda L^2=0,$ has as members $S=0$ for $\lambda=0$ and the double line $L^2=0,$ for $\lambda = \infty.$ ($\lambda$ is usually thought of as belonging to ${\Bbb P}^1$ and one sees $\mu S+\lambda L^2=0,$ where $\lambda, \mu$ are homogeneous coordinates on ${\Bbb P}^1$ as an alternate form of the same pencil).
Every other member in the pencil will be tangent to $S=0$ at the two points of intersection $S=0\cap L=0.$ The fixed locus of the pencil are the two double points of this intersection.
Aside Enumerative geometry is based on notions like "touching at these two points" being preserved in families. A curve intersecting a double line technically is tangent there; this is why $6^5=7776$ (A naive application of Bezout's theorem on the fact that the condition to be tangent to a conic is of degree 6) is the wrong answer to the question "How many conics are tangent to 5 given conics," the double lines in some precise sense contributes the difference to the actual number 3264.
The problem to find the parabola $P$ in this family can be solved by looking for the $\lambda$ for which $S+\lambda L^2=0$ has a quadratic term that is a square of linear terms, this is what the discriminant gives us: $B^2-4AC=(2\lambda+1)^2-4(\lambda+1)^2=0,$ which gives $\lambda=-\frac34,$ or $P=\frac14(x^2-2xy+y^2-2x-2y+1)=0$ or $P=\frac14((x-y)^2-2x-2y+1)=0.$