Tangent to ellipse rotated in 3D with perspective

Question

I have an ellipse in 3D, with width $a$, height $b$, centered on $(0, 0, C)$ and parallel to the $xy$-plane.

I'm using a parametric equation with parameter $t$:

$x = a \cdot cos(t) \\ y = b \cdot sin(t) \\ z = C$

I rotate it by $\theta$ around the $z$-axis and then by $\phi$ around the $x$-axis to get:

$x_r = x \cdot cos(\theta) - y \cdot sin(\theta) \\ y_r = (x \cdot sin(\theta) + y \cdot cos(\theta)) \cdot cos(\phi) - z \cdot sin(\phi) \\ z_r = (x \cdot sin(\theta) + y \cdot cos(\theta)) \cdot sin(\phi) + z \cdot cos(\phi)$

Then I project it on the $xy$-plane with a perspective transform using some constant $p$:

$x_p = \dfrac{p \cdot x_r}{p + z_r} \\ \\ y_p = \dfrac{p \cdot y_r}{p + z_r}$

Before applying perspective, I can calculate the tangent (in 2D) with:

$dx = x \cdot cos(\theta) + y \cdot sin(\theta) \\ \\ dy = (x \cdot sin(\theta) - y \cdot cos(\theta)) \cdot sin(\phi)$

How can I get an equation for the 2D tangent in terms of $t$ for the ellipse after the perspective transform has been applied?

Answer 1

(Disclaimer: I have never worked with a perspective transformation before, and it's been a while since I've thought about rotating things in 3D space. But using the formulas provided in the question, I think I can provide an answer using basic principles of calculus.)

If we want to compute the slope of the tangent line of the projected curve given parametrically by \begin{align} x_p(t) &= \frac{p\,x_r(t)}{p+z_r(t)} \\ y_p(t) &= \frac{p\,y_r(t)}{p+z_r(t)} \end{align} we can do it by computing $dx_p/dt$ and $dy_p/dt$ and then using the fact that $$ \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} $$ How do we compute those two derivatives? By a long application of the chain rule. I think trying to write out the entire expression for the derivatives would be unwieldy. Therefore I will not expand the new factors the chain rule gives us, but will provide the formulas to compute them. I don't know what the application of this question is, but if you'll be coding it up, this is probably the better way to do it.

Let us begin. First I'll compute $dx_p/dt$:$\newcommand{\indent}{\ \ \ \ \ \ }$ \begin{align} \frac{dx_p}{dt} &= \frac{d}{dt} \left(\frac{px_r}{p+z_r} \right) \\ &= \frac{p\,x_r' \cdot (p+z_r) - px_r z_r'}{(p+z_r)^2} \indent \textrm{Quotient Rule} \end{align} where \begin{align} x_r' &= x' \cos \theta - y' \sin \theta \\ z_r' &= (x' \sin \theta + y' \cos \theta) \sin \phi \end{align} and \begin{align} x' &= -a \sin t \\ y' &= b \cos t \end{align} And $dy_p/dt$ looks the same just with $y_r$ instead of $x_r$: $$ \frac{dy_p}{dt} = \frac{p\,y_r' \cdot (p+z_r) - py_r z_r'}{(p+z_r)^2} $$ where $$ y_r' = (x' \sin \theta + y' \cos \theta) \cos \phi $$ If we go ahead and take the quotient of these two, we get a small simplification since their common denominators of $(p+z_r)^2$ will cancel: \begin{align} \boxed{\frac{dy_p}{dx_p} = \frac{py_r' \cdot (p+z_r)-py_r z_r'}{px_r' \cdot (p+z_r)-px_r z_r'}} \end{align} It may be that this can be further simplified if you plug in the corresponding expressions for the various variables in this expression, but from a cursory inspection, it doesn't look like it will.