# Perspective Ellipse Math for SoK'22

Tymon Dąbrowski tamtamy.tymona at gmail.com
Mon Mar 14 07:00:00 GMT 2022

```Hi @Srirupa,

***Method 1***
I checked your PDF and in my opinion, the result you got from Wolfram Alpha
is consistent with your data and your approach. If you use the values for
a, b, c,..., into the generic ellipse equation, you'll see that it results
in an equation:
a*x^2 + a*y^2 - a = 0
which is the same as:
a*(x^2 + y^2 - 1) = 0
And you can see that no matter what value "a" is, it represents the same
points on the plane (except when a is equal to 0, then it represents the
whole plane and not just the ellipse).

The problem in your equation is that every part of the equation has a
parameter/variable next to it. It's probably easiest to see if you reduce
by *f *(here, with the assumption that *f* != 0):
f * (a/f * x^2 + b/f * xy + c/f * y^2 + d/f * x + e/f * y + 1) = 0
as you can see, no matter what value f is, it will represent the exact same
ellipse. So you could represent it this way:
a/f * x^2 + b/f * xy + c/f * y^2 + d/f * x + e/f * y + 1 = 0
or, since *a/f* is just as unknown as *a*, this way:
a * x^2 + b * xy + c * y^2 + d * x + e * y + 1 = 0
and this reduces the dimension of the linear equations system to just 5.

NOW, I'm not 100% sure if you can just reduce *f* like that, because I'm
not sure whether *f* (in the original equation) can or cannot be 0. I
haven't explored that possibility before writing this mail. If* f* *can* be
0, it would have to be considered somewhere separately.
Thinking about it, while I'm not sure whether we can reduce *f*, we
*probably* could reduce for example *a*.
I'm not 100% sure about it either, you'd have to check that, but I'm
somewhat quite sure you couldn't have an ellipse with equation like that:
0*x^2 + ... = 0
so I think it would be fine if you reduce by *a* and change the original
equation to be:
a * (1 * x^2 + b/a * xy + c/a * y^2 + d/a * x + e/a * y + f/a) = 0
which means you could make an equation like that:

*1 * x^2 + a * xy + b * y^2 + c * x + d * y + e = 0*

which also means just 5 variables/equations, not 6, and less chance of
infinite solutions.

Think about it this way: it's similar to as if you had an equation for a
line looking like this: "ax + by + c = 0", which looks correct, but then
you'd try having three equations to find all three a, b and c, while the
line only needs two points to be defined, which means just two equations.
To represent a line, at least one of a, b, or c must not be 0, so there is
always a way to reduce the number of equations to 2 (though it depends on
the data in that case because sometimes you'd have "y = 0" (so a = 0, b =
1, c = 0) and sometimes you'd have "x = 0" (a = 1, b = 0, c = 0)).

-----

Also
> I need to solve a system of linear equations for six variables and this
is *probably* complicated.
Actually, you'd probably just have to put it all into a matrix and a vector
and ask Eigen to solve it for you. It's a very basic operation (it's not
trivial to write yourself, though it wouldn't be very difficult either, but
it's so common of a need that I would be surprised to learn that some
library offering matrix operations doesn't provide it out of the box. As an
example, in languages "octave" (foss) and "matlab" (proprietary) you can
solve a system of linear equations just by writing a few characters: "x =
A\b" (note the backslash instead of slash) when A is the matrix of
coefficients and b is the vector of constant terms. And x is the solution
vector. The original equation system is Ax = b).
In Eigen you'll probably have to write a few more characters than that, but
it should be fairly easy, too :)
For Eigen, here you have documentation that explains what needs to be
written: https://eigen.tuxfamily.org/dox/group__TutorialLinearAlgebra.html
And for usage of Eigen in Krita, you can refer to the file:
libs/global/kis_algebra_2d.cpp

***Method 2***
I think there should be a way, but I don't know yet, but also you have
Method 1 already figured out, so maybe it's not necessarily needed.

---

Hope that helps,
Tiar

PS. If anything is unclear, feel free to ask, either here or on IRC (tiar).

niedz., 13 mar 2022 o 19:04 Srirupa Datta <srirupa.sps at gmail.com>
napisał(a):

> I have a quadrilateral, whose 4 corners are known, and an inscribed
> ellipse whose touching points with the quadrilateral and also known. I need
> to find out the extreme points of the major axis of that ellipse.
>
> *Method 1*
> I decided to first find out the equation of the inscribed ellipse. This
> is what I did so far. However this approach has a challenge. I need to
> solve a system of linear equations for six variables and this is
> *probably* complicated. So I'm not sure if this is at all the right
> approach.
>
> Also, I got stuck since I expected the solution to be a unique one.
>
> *Method 2*
> I have the transformation matrix that maps a unit square to my
> quadrilateral. Is there any way I can use transformation mathematics to
> solve this problem easily?
>
> Thanks,
> Srirupa
>
>
>
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