Inscribing an ellipse in an irregular convex pentagon Using the methods of projective geometry, identify the unique ellipse that is inscribed in a given convex pentagon.
Suppose the vertices of the pentagon are: $(1, 0), (4, 2), (3, 6), (-1, 5), (-1, 1)$.  Find the equation of the unique inscribed ellipse that is tangent to all five sides of this convex pentagon.


It's easier to find the dual conic first.
Rewrite every tangents in the form:
$$X_i x+Y_i y+1=0$$
Then $(X_i,Y_i)$ are the five points defining the dual conic:
$$
\begin{align}
  0 &= \det
  \begin{pmatrix}
    X^2 & XY & Y^2 & X & Y & 1 \\
    X_1^2 & X_1 Y_1 & Y_1^2 & X_1 & Y_1 & 1 \\
    \vdots & & & & & \vdots \\
    X_5^2 & X_5 Y_5 & Y_5^2 & X_5 & Y_5 & 1 \\
  \end{pmatrix} \\ \\
  &= AX^2+2HXY+BY^2+2GX+2FY+C \\
\end{align}$$
The required ellipse is
$$
\begin{align}
  0 &= -\det
  \begin{pmatrix}
    0 & x & y & 1 \\
    x & A & H & G \\
    y & H & B & F \\
    1 & G & F & C \\
  \end{pmatrix} \\ \\
  &= ax^2+2hxy+by^2+2gx+2fy+c
\end{align}$$
Note that $a$ is the co-factor of entry $A$ of the $3\times 3$ block matrix, etc.
For the vertical tangent, try to let $Y_i=N$ and only the terms with highest order in $N$ survive.
See also the case of quadrilateral here.
