Prove that perspective projection of circle is ellipse How to prove that perspective projection of a circle is an ellipse?
I start with the parametric equation of circle and ellipse:
Circle:
$x = r\cos t$
$y = r\sin t$
Ellipse:
$x = a\cos(t)$
$y = b\sin(t)$
Then I have perspective projection matrix:
$$  
    \begin{pmatrix}
    a_{11} & a_{12} & a_{13} \\
    a_{21} & a_{22} & a_{23} \\
    a_{31} & a_{32} & 1 \\
    \end{pmatrix}
$$
And obtain projected coordinates (affine transform for simplicity):
$$  
    \begin{pmatrix}
    a_{11} & a_{12} & a_{13} \\
    a_{21} & a_{22} & a_{23} \\
    0 & 0 & 1 \\
    \end{pmatrix}
    \begin{pmatrix}
    x \\
    y \\
    1 \\
    \end{pmatrix}
    =
    \begin{pmatrix}
    u \\
    v \\
    1 \\
    \end{pmatrix}
$$
If we convert this to equations:
$u = a_{11}x+a_{12}y+a_{13}$
$v = a_{21}x+a_{22}y+a_{23}$
Substituting $x,y$ with parametric equations:
$a\cos t = a_{11}\,r\, \cos t + a_{12}\, r \,\sin t + a_{13}$
$b \sin t = a_{21}\,r \,\cos t + a_{22}\, r \,\sin t + a_{23}$
What to do next? I can't see how these equations can be equal.
 A: As noted in the answer of @rfabbri ( +1), the perspective projection of a circle is not always an ellipse, but it is in general a conic section: an ellipse, a  parabola, or a hyperbola.
If you want a matrix that transform the circle in an ellipse by projection, than note that the general equation of a conic:
$$
Ax^2+Bxy+Cy^2+Dx+Ey+F=0
$$
can be written in the form (see here):
$$
\begin{pmatrix}
x&y&1
\end{pmatrix}
\begin{pmatrix}
A&\frac{B}{2}&\frac{D}{2}\\
\frac{B}{2}&C&\frac{E}{2}\\
\frac{D}{2}&\frac{E}{2}&F\\
\end{pmatrix}
\begin{pmatrix}
x\\y\\1
\end{pmatrix}=0
$$
so, for $A=C=1$ and $B,D,E=0$ and $F=-r^2$ we have the circle center at the origin and radius $r$, that becomes  an ellipse if the matrix is such that
$B^2-4AC <0$ and the conic is not degenerate, i.e. the determinat of the matrix is not null.
A: You cannot, because the perspective projection of a circle is not always an ellipse. Draw a cone where the tip of the cone is the center of projection and the cone goes through the circle. It is easy to imagine a configuration of the projection plane that cuts the cone into a curve (a conic section) that is not an ellipse.
A: Your last step is wrong, you shouldn't substitute the original coordinates in the LHS. On the opposite, consider that you have a curve of parameteric equations
$$\begin{cases}x=a\cos t+b\sin t,\\y=c\cos t+d\sin t.\\\end{cases}$$ (You can translate the origin to absorb the constant terms.)
Inverting this sytem, you will get 
$$\begin{cases}\cos t=px+qy,\\\sin t=rx+sy\\\end{cases}$$ which yields the implicit equation of a conic,
$$(px+qy)^2+(rx+sy)^2=1.$$
