Our teacher has given us a question to be solved.

What curve does the intersection points of the given lines make? A parabola, hyperbola, or none of them? (please look at the image I've posted. I am not speaking about all of the intersection points. Just the tangent curve beneath the lines)

the lines are as follows:












my solution: I tried to find if it is a parabola first. Assuming the maximum point to be $(0,\frac{13\sqrt{3}}{5})$ (the intersection point of $4$th and $6$th lines), I found the equation to be $y=-\frac{13\sqrt{3}}{125}x^2+\frac{13\sqrt{3}}{5}$. But after plotting it in desmos, the curve didn't meet the intersection points.

Then I tried to assume $(0,\frac{5\sqrt{3}}{2})$ as the maximum; I yielded $y=-\frac{\sqrt{3}}{10}x^2+\frac{5\sqrt{3}}{2}$; but it again failed in plotting.

I then switched to hyperbola: After a long effort I found the equations to be $$\frac{(y-5\sqrt{3})^2}{(\frac{5\sqrt{3}}{2})^2}-\frac{x^2}{(\frac{5\sqrt{3}}{3})^2}=1$$ and $$\frac{(y-5\sqrt{3})^2}{(\frac{12\sqrt{3}}{5})^2}-\frac{x^2}{(\frac{5\sqrt{3}}{3})^2}=1$$ but I hink both of them are incorrect; because they are irrelevant to their asymptotes ($10$th & $11$th lines in the set of lines).

line sets image

Sorry, I had to stretch the picture a little so that it better be displayed.

  • $\begingroup$ I am still not sure if those lines are asymptotes. $\endgroup$
    – user815214
    Aug 11, 2020 at 14:00
  • $\begingroup$ Can you/your teacher be more specific about which intersection points between which lines? If you have 11 non-parallel lines that gives you 110 different intersection points, which you can see on the graph. Doesn't make sense to talk about a degree 2 conic equation that satisfies 110 intersection points... $\endgroup$
    – user810677
    Aug 11, 2020 at 14:01
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    $\begingroup$ Once you've decided which points are the relevant ones - I would just use the fact that all conics are uniquely defined by 5 points. Or choose any 6 points and sub them into ax^2 + by^2 + cxy + dx + ey + f = 0 and solve the six linear equations for a, b, c, d, e, f. $\endgroup$
    – user810677
    Aug 11, 2020 at 14:06
  • 3
    $\begingroup$ @MathGeek: It appears that you seek the curve that is the so-called "envelope" of the family of lines. An envelope isn't defined to go through intersection points of the lines; rather it's tangent to each of the lines, which agrees with this notion that it's the curve you "see" as more and more of those lines are drawn. $\endgroup$
    – Blue
    Aug 11, 2020 at 14:17
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    $\begingroup$ @MathGeek: I agree that the Wikipedia entry is a bit opaque. A site search here for "envelope" may yield clear examples. That said, finding an envelope typically requires a formula for an infinite family of tangent lines, to which we apply some Calculus (which you may not be able to use). A mere finite list of lines only makes the problem harder and just-plain tedious. I'd have to ponder about the approach your instructor intends; maybe there's a neat trick here. ... Could it be that you aren't looking for the eqn at all, but are just supposed to graph and exclaim "Hey, neat! A parabola!"? $\endgroup$
    – Blue
    Aug 11, 2020 at 14:49

1 Answer 1


Inspection shows that the given lines can be produced as follows: $$y_k(x)={\sqrt{3}\over10}\bigl(2kx+(k^2+25)\bigr)\qquad(-5\leq k\leq 5)\ .\tag{1}$$One now should look for the envelope of the family $\bigl(\ell_c\bigr)_{c\in{\mathbb R}}$ of lines $$\ell_c:\quad y-y_c(x)=0\ ,$$where the integer $k$ in $(1)$ has been replaced by the real parameter variable $c$.

Doing the standard calculations for finding the envelope one finds that the envelope $\epsilon$ of the line family $\bigl(\ell_c\bigr)_{c\in{\mathbb R}}$ is the parabola $$\epsilon:\qquad y={\sqrt{3}\over10}(25-x^2)\ ,$$ plotted in red in the following figure.

Referring to these "standard calculations": When a family of curves in the $(x,y)$-plane is given by an equation of the form $$F(x,y,c):=y-{\sqrt{3}\over10}\bigl(2c x+(c^2+25)\bigr)=0$$ then most points $(x,y)$ are "ordinary" points, meaning that there is no curve going through them, or that in the neighborhood of $(x,y)$ the curves are lined up like nice "parallel" stream lines. When $$F(x_0,y_0,c_0)=0\qquad\wedge\qquad F_c(x_0,y_0,c_0)\ne0$$ then $(x_0,y_0)$ is an "ordinary" point. When $$F(x_0,y_0,c_0)=0\qquad\wedge\qquad F_c(x_0,y_0,c_0)=0\tag{1}$$ then $(x_0,y_0)$ is a "special" point, since $c$, defined by $F(x,y,c)=0$, is no longer a good function of $(x,y)$ in the neighborhood of $(x_0,y_0)$.

Some of the "special" points are indeed lying on an envelope $\epsilon$ of the given curve family. You obtain these "special" points by eliminating $c$ from the two equations $(1)$.

enter image description here

  • $\begingroup$ “Doing the calculations” — What calculations? $\endgroup$ Aug 11, 2020 at 19:12
  • $\begingroup$ Please explain it. how did you conclude the envelope is a parabola with that function? With which calculations? in Envelope Wikipedia article I have found calculations for finding the envelope curve (starting with the equation: $$\frac{x}{t}+\frac{y}{11-t}=1$$ $11$ is just an example $\endgroup$
    – user815214
    Aug 12, 2020 at 8:51

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