# Understanding Conics in Pencil

http://planetmath.org/pencilofconics

I found some simple examples and thought that maybe in the case of ellipses I understood enough to keep reading, but now I'm not so sure. From what I understand from the planetmath website is that any two conics (in my case ellipses) $H_1$ and $H_2$ generate a pencil by their four intersection points, that then generate four lines $L_1$, $L_2$, $L_3$, and $L_4$ and the pencil is $pL_1L_2=qL_3L_4$ where $p$ and $q$ are free parameters. All of this I feel comfortable with, until the paper says the following (paraphrasing a bit):

"... positions 10, 9, and 7 are separated according to the nature of the degenerate conics in the pencil $\alpha A+B$ in the following way:
position 10: if there are no degenerate conics.
position 9: if they are a pair of real lines.
position 7: if the only degenerate conic in the pencil is a double real line."

Where $\alpha$ is a scalar and $A$ and $B$ are two $3\times3$ matrix representation of two different ellipses. I don't understand how once we make a new $3\times3$ matrix $C=\alpha A+B$ that corresponds to a pencil? I though we needed two conics to make a pencil and so we would need another $3\times3$ matrix besides $C$ to generate a pencil. If I'm not being clear then please let me know I can always elaborate. Or if it's clear that I just need to read more I would be grateful for resources. Also, the paper I'm reading can be found here:

http://www.sciencedirect.com/science/article/pii/S0167839606000033

• I think the idea is that $\alpha$ is a variable. By setting it to different values you get all the conics in the pencil. – Nefertiti Jul 21 '16 at 20:04
• @Nefertiti Thanks so much for replying!! In the paper they have $\alpha = \frac{-a}{3}$, where a is the coefficient for $\lambda^2$ in the polynomial $f(\lambda) = det( \lambda A + B)$ where A and B are again the 3x3 matrix representations for each ellipse respectively. So it looks as though $\alpha$ is fixed, but what you said makes sense to me.. How would one go about testing whether $\alpha A + B$ is ever a degenerate conic, as in position 10? – mike van der naald Jul 21 '16 at 20:59
• OK, I looked at the paper. I think the passage you are asking about is very confusingly written. What they appear to be saying is that for this specific value, $\alpha A + B$ is the (unique) degenerate conic in the pencil. So the plural "conics" in the first line and the pronoun "they" later on are incorrect. So to sum up: 1) the pencil is $\lambda A +B$ where $\lambda$ is a variable. 2) $\alpha A +B$ is a specific conic in the pencil, the unique degenerate one. 3) the referee did a bad job. – Nefertiti Jul 22 '16 at 8:45
• @Nefertiti Thank you so much for taking your time to look at the paper. I believe you are correct, the three cases that follow make a lot more sense if you call $\alpha A + B$ the only degenerate conic. I have emailed the author to try and clarify but I cannot thank you enough for helping me out here. If you want to type this comment up as an answer (which I feel is more than appropriate) I would be happy to accept it. – mike van der naald Jul 22 '16 at 16:29
• OK, I made my comment into an answer. Glad to be of help! – Nefertiti Jul 22 '16 at 18:43

OK, I looked at the paper. I think the passage you are asking about is very confusingly written. What they appear to be saying is that for this specific value, $\alpha A +B$ is the (unique) degenerate conic in the pencil. So the plural "conics" in the first line and the pronoun "they" later on are incorrect. So to sum up: 1) the pencil is $\lambda A +B$ where $\lambda$ is a variable. 2) $\alpha A +B$ is a specific conic in the pencil, the unique degenerate one. 3) the referee did a bad job.
$C$ is made from the two matrices $A$ and $B$ each defining a conic. Let $\alpha=0$ and you get the one, let $\alpha=\infty$ and you get the other. The whole pencil is described by varying $\alpha$. (If you don't feel comfortable with one conic appearing only in the limit $\alpha\to\infty$, consider $\lambda A+(1-\lambda) B$ instead.)
Assuming the conics meet in four points, $Ap_k=Bp_k=0$ so that $(\alpha+B)p_k=0$ and all the conics in the pencil have these points in common. In particular, there are conics in the pencil that are degenerate, i.e. pairs of lines through the four points.
• Thanks for you comment! In the paper linked they let $\alpha = \frac{-a}{3}$ where a is the coefficient for $\lambda ^2$ in the polynomial $f( \lambda ) = det( \lambda A + B)$, so it appears that $\alpha$ is fixed? Maybe I'm not reading the paper correctly, though. – mike van der naald Jul 21 '16 at 21:20
• Fixing $\alpha$ selects a particular conic out of the pencil. – Yves Daoust Jul 21 '16 at 21:25
• This is where my confusion comes in. The author fixes alpha and then begins talking about conics inside $\alpha A + B$. For example: position 10, they say " if there are no degenerate conics in the pencil." which is confusing since they've selected a single conic out of the pencil. I guess I'm not sure if I'm not reading the author correctly, or if there is a typo and the three cases he describes are actually talking about one conic that is $\alpha A + B$? This would make sense seeing how "position 9: if they are a pair of real lines." doesn't seem to be a statement about pencils. – mike van der naald Jul 21 '16 at 21:31