Prove that perspective projection of circle is ellipse

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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,

$$(px+qy)^2+(rx+sy)^2=1.$$