The Wikipedia article on fashionable triangle geometry has a picture labeled “Artzt parabolas” with no rationalization.
A fast search didn’t flip up something about Artzt parabolas [1], however apparently the parabolas undergo pairs of vertices with tangents parallel to the perimeters.
The final type of a conic part is
ax² + bxy + cy² + dx + ey + f = 0
and the constraint b² = 4ac means the conic shall be a parabola.
We have now 6 parameters, every decided solely as much as a scaling issue; you’ll be able to multiply either side by any non-zero fixed and nonetheless have the identical conic. So a normal conic has 5 levels of freedom, and the parabola situation b² = 4ac takes us all the way down to 4. Specifying two factors that the parabola passes by takes up 2 extra levels of freedom, and specifying the slopes takes up the final two. So it’s believable that there’s a distinctive resolution to the issue.
There may be certainly an answer, distinctive as much as scaling the parameters. The next code finds parameters of a parabola that passes by (xi, yi) with slope mi for i = 1, 2.
def clear up(x1, y1, m1, x2, y2, m2):
Δx = x2 - x1
Δy = y2 - y1
λ = 4*(Δx*m1 - Δy)*(Δx*m2 - Δy)/(m1 - m2)**2
okay = x2*y1 - x1*y2
a = Δy**2 + λ*m1*m2
b = -2*Δx*Δy - λ*(m1 + m2)
c = Δx**2 + λ
d = 2*okay*Δy + λ*(m1*y2 + m2*y1 - m1*m2*(x1 + x2))
e = -2*okay*Δx + λ*(m1*x1 + m2*x2 - y1 - y2)
f = okay**2 + λ*(m1*x1 - y1)*(m2*x2 - y2)
return (a, b, c, d, e, f)
[1] The web page stated “Artz” after I first checked out it, nevertheless it has since been corrected to “Artzt”. Perhaps I didn’t discover something as a result of I used to be in search of the mistaken spelling.
