Families of Conics

What this move does

If a point $P$ satisfies both $S_1=0$ and $S_2=0$, then for every $\lambda$ the value of $S_1+\lambda S_2$ at $P$ is $0+\lambda \cdot 0=0$:

A conic is determined by $5$ conditions. The common points use up $4$ of them, leaving one parameter $\lambda$:

1. Write the family: find two curves $S_1$, $S_2$ that are easy to write down and both pass through these points (often products of lines).
2. Fix the parameter: use the remaining condition (one more point, being a parabola, being a circle…) to get an equation for $\lambda$.
3. Check for gaps: do not forget $S_2=0$ itself ($\lambda =\infty$).

For four points $A,B,C,D$, the handiest pair is “the two pairs of opposite sides”: $S_1=AB\cdot CD$, $S_2=BC\cdot DA$ (each symbol stands for the left side of that line's equation). Two circles work too: subtracting one circle from the other ($\lambda =-1$) gives the line of their common chord.

Work through one

When it fails

The family only guarantees “passes through the common points”. If two curves have fewer than four real common points (for example, two circles meeting in only two points), $S_1+\lambda S_2=0$ still passes through those real points, but it is no longer “all” the conics through them: conics through two points still have three degrees of freedom. For two circles, $S_1+\lambda S_2=0$ ($\lambda \ne -1$) gives exactly all the circles through their intersection points, because “being a circle” uses up two more conditions.

Practice