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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.

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    $\begingroup$ Please write a_{11} for indexed variables. You could also consider removing the multiplication operator * because in modern mathematics this operator is implicitly implied if the variables are written without an operator in the middle of them. If the spacing is to small you can use \, or \quad \qquad for larger spacings between the variables. $\endgroup$
    – MrYouMath
    Mar 3, 2018 at 14:24
  • $\begingroup$ You’re assuming that the center of the ellipse will coincide with the center of the circle at the origin. Why do you believe that? That problem aside, you can always rewrite the sum of two sinusoids that have the same period as a single sinusoidal term. $\endgroup$
    – amd
    Mar 3, 2018 at 22:43

3 Answers 3

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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 (maybe degenerate).

If you want a matrix that transform the circle in an ellipse by projection, than note that teh 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.

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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.

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    $\begingroup$ This answer feels a bit misleading to me. As long as the circle isn't infinitely far away and also doesn't touch or cross the plane containing the center of projection that is tangent to the view direction, then the perspective projection of a circle is an ellipse. The question should probably mention that, but it'd maybe be better to say that the proof is possible with extra constraints, rather than to say it's not possible? The proof of why the plane-cone intersection is an ellipse whenever it's a closed curve is interesting. Wouldn't that make a better answer than "you cannot"? $\endgroup$
    – David
    Nov 18, 2018 at 3:46
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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,

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

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