I graphed a square with sides:
$y=x$
$y=-x+40$
$y=x+40$
$y=-x$
Then I divided each side to 20 equal parts and drew lines through points:
$(1,39)$ & $(19,19)$: (line 1)
$(2,38)$ & $(18,18)$: (line 2)
etc.
Then I found the equation of the envelope of the set of lines by finding 5 intersection points (between lines 1 & 2, 2 & 3, . . .) and plugging them in the formula: $$\frac{A}{F}x^2+\frac{B}{F}xy+\frac{C}{F}y^2+\frac{D}{F}x+\frac{E}{F}y+1=0$$
After simplifications I found the envelope equation to be: $$-y^2-40x+40y+1=0$$
Which is obviously a parabola.
But then I decided to find the equation of envelope of the set of lines in another two sides of the square:
line 1: through points $(-1,39)$ & $(19,21)$
line 2: through points $(-2,38)$ & $(18,22)$
etc.
Again I found the intersection points of the lines 1 & 2, 2 & 3, . . . and placed them in above formula.
After simplifications I found the envelope equation to be: $$178749x^2+165738xy+138115y^2-5552223x-4092601y+36030=0$$
Which is an ellipse!
Why do the conic sections differ while both the envelopes are the same and they are just rotated in resepct to each other?
Usually you would define the envelope not as the conic passing through 5 intersection points, but tangent to five of the lines. So you would compute the dual conic first (i.e. the conic defined as a set of tangent lines), and then the primal from that (i.e. the conic as a set of incident points).
Doing that for your first set you get the dual conic as
$$(a,b,c)\cdot\begin{pmatrix}20&20&1\\20&20&0\\1&0&0\end{pmatrix}\cdot\begin{pmatrix}a\\b\\c\end{pmatrix}=0$$
so a line $ax+by+c=0$ is tangent to the conic if it satisfies the quadratic equation above. The primal conic is then any multiple of the inverse of that matrix, so you get that as
$$(x,y,1)\cdot\begin{pmatrix}0&0&20\\0&1&-20\\20&-20&0\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix}=0$$
which is a parabola since the determinant of the top left $2\times 2$ matrix is zero. You may write this as $-y^2-40x+40y=0$ which is close to the $-y^2-40x+40y+1=0$ you have in your question but not quite the same as the constant term differs.
For your second set of lines you get
$$(a,b,c)\cdot\begin{pmatrix}20&0&0\\0&60&1\\0&1&0\end{pmatrix}\cdot\begin{pmatrix}a\\b\\c\end{pmatrix}=0 \\ (x,y,1)\cdot\begin{pmatrix}1&0&0\\0&0&20\\0&20&-1200\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix}=0$$
which is again a parabola. You might also write it as $x^2=1200-40y$ if you prefer, which is an equation already discussed in the comments. Or as $y=30-\frac1{40}x^2$ which may be even more familiar as the equation of a parabola.
But even if using the tangents instead of intersections is the correct way of getting the envelope correctly without any approximation, if you were using intersections in a symmetrical way you should get a symmetrical result. So let's check. Here are the first 5 points of intersections of consecutive lines for your first set (the decimal numbers are exact, not rounded):
$$ \begin{pmatrix}2.8\\37\end{pmatrix}\quad \begin{pmatrix}4.4\\35\end{pmatrix}\quad \begin{pmatrix}5.8\\33\end{pmatrix}\quad \begin{pmatrix}7.0\\31\end{pmatrix}\quad \begin{pmatrix}8.0\\29\end{pmatrix}\\ (x,y,1)\cdot\begin{pmatrix}0&0&20\\0&1&-20\\20&-20&-1\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix}=0 $$
So your $-y^2-40x+40y+1=0$ with its constant term is correct here. For the second set of lines you get
$$ \begin{pmatrix}17\\22.8\end{pmatrix}\quad \begin{pmatrix}15\\24.4\end{pmatrix}\quad \begin{pmatrix}13\\25.8\end{pmatrix}\quad \begin{pmatrix}11\\27.0\end{pmatrix}\quad \begin{pmatrix}9\\28.0\end{pmatrix}\\ (x,y,1)\cdot\begin{pmatrix}1&0&0\\0&0&20\\0&20&-1201\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix}=0 $$
Again a parabola. Again almost the same as what you get from the tangents, except for a shift by $1$ in the constant term. And again an equation which has already been given, namely as $x^2 = -40y + 1201$ in one of the comments.
I don't know where you got that equation of an ellipse from, but I suggest you check the computation leading to that. Probably no conceptual problem there but just a typo somewhere or something like that.