What kind of curve do these lines make? 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:

$y=\frac{4\sqrt{3}}{5}x+\frac{41\sqrt{3}}{10}$
$y=\frac{3\sqrt{3}}{5}x+\frac{17\sqrt{3}}{5}$
$y=\frac{2\sqrt{3}}{5}x+\frac{29\sqrt{3}}{10}$
$y=\frac{\sqrt{3}}{5}x+\frac{13\sqrt{3}}{5}$
$y=\frac{5\sqrt{3}}{2}$
$y=-\frac{\sqrt{3}}{5}x+\frac{13\sqrt{3}}{5}$
$y=-\frac{2\sqrt{3}}{5}x+\frac{29\sqrt{3}}{10}$
$y=-\frac{3\sqrt{3}}{5}x+\frac{17\sqrt{3}}{5}$
$y=-\frac{4\sqrt{3}}{5}x+\frac{41\sqrt{3}}{10}$
$y=\sqrt{3}x+5\sqrt{3}$
$y=-\sqrt{3}x+5\sqrt{3}$

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.

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$$\ell_c:\quad y-y_c(x)=0\ ,$$where the integer $k$ in $(1)$ has been replaced by the real parameter variable $c$.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 $(1)$.

