Origin-Centred Elliptical "Spotlight" with Conical Light Source of Fixed Aperture 
You have a light source with conical semi-aperture angle of $u$, and you want to create an ellipse-shaped spotlight with equation $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$ $(a>b)$ by shining it on the Cartesian $x$-$y$ plane from $(k,0,h)$ at a tilt angle $\alpha$ to the vertical on the $x$-$z$ plane. Express $k,h,\alpha$ in terms of $u,a,b$.

In other words, using a light source with a fixed aperture, where should you place it on the $x$-$z$ plane and at what angle from the vertical should you tilt it from  in order to create an ellipse of given semi-major and semi-minor axes centred at the origin?
(NB - this is a variation of this other question here posted earlier)
Here's a nice video by ElicaTeam illustrating something similar. A screenshot is shown below.

A screenshot from a desmos simulation is shown below. 
$\hspace{2cm}$]2

Addendum
The answer that I've worked out is 
$$\boxed{\begin{align}
&\sin\alpha&&=\sqrt{1-\frac{b^2}{a^2}} \cos u&&=e \cos u\\
&k&&=\sqrt{\left(1-\frac{b^2}{a^2}\right)(a^2+b^2\cot^2u)}&&=e\sqrt{a^2+b^2\cot^2u}\\
&h&&=\frac {b^2}{a\tan u}&&=a(1-e^2)\cot u\end{align}}$$
where $e=\sqrt{1-\frac{b^2}{a^2}}$ is the eccentricity of the ellipse. 
See desmos implementation here.
Interestingly, there are similarities with the tilted martini glass problem posted here on MSE last year, and originally posted on $538$ Riddler. 

Further Addendum (27 Jul 2017)
See improved desmos implementation here. The red curve is the locus of the light source position for constant $a$ and the blue curve is the locus for constant $b$. 

 A: Let $A=(a,0,0)$ and $B=(-a,0,0)$ be the endpoints of the ellipse major axis, and $V=(k,0,h)$ the cone vertex, with $\angle AVB=2u$. Let then $P$ be any point on the ellipse and $H$ its projection onto $AB$. A plane through $P$, perpendicular to the axis of the cone, intersects the cone along a circle $A'B'P$ (see picture below), where $A'$ and $B'$ lie on plane $VAB$.
By the intersecting chords theorem we know that $PH^2=A'H\cdot B'H$. But, on the other hand, we have by similitude: 
$$
A'H:AH=BC:AB,
\quad\hbox{that is:}\quad
A'H={BC\over AB}\cdot AH;
$$
$$
B'H:BH=AD:AB,
\quad\hbox{that is:}\quad
B'H={AD\over AB}\cdot BH.
$$
If we set $n=VA$ and $m=VB$, the above formulas can be written as 
$$
A'H={m\sin u\over a}\cdot AH,
\quad
B'H={n\sin u\over a}\cdot BH,
$$
and inserting these into the formula for $PH^2$ we get:
$$
PH^2={mn\sin^2 u\over a^2}\,AH\cdot BH.
$$
When $H$ is the midpoint of $AB$, then $PH=b$ and $AH=BH=a$, and from the above equation we get:
$$
mn={b^2\over \sin^2 u}.
$$
Another equation for $m$ and $n$ can be found from the cosine rule applied to triangle $AVB$:
$$
m^2+n^2-2mn\cos2u=4a^2.
$$

From the above equations you can find $m$ and $n$ in terms of $u$, $a$, $b$:
$$
{m^2+n^2\over2}=2(a^2-b^2)+{b^2\over\sin^2u},
\quad
{m^2-n^2\over2}={2\over\sin u}
\sqrt{(a^2-b^2)(a^2\sin^2u+b^2\cos^2u)}.
$$
From $m$ and $n$ it then easy to find $\alpha$, $h$ and $k$: if $M$ is the intersection between $AB$ and the axis of the cone, we have:
$$
\angle VMB={\pi\over2}+\alpha,\quad
\angle VMA={\pi\over2}-\alpha,\quad
\angle VBM={\pi\over2}-u-\alpha,\quad
\angle VBM={\pi\over2}-u+\alpha,
$$
and from the sine rule:
$$
{VM\over\cos(u-\alpha)}={n\over\cos\alpha},
\quad
{VM\over\cos(u+\alpha)}={m\over\cos\alpha},
\quad\hbox{whence:}\quad
{\cos(u-\alpha)\over\cos(u+\alpha)}={m\over n}.
$$
We can solve the last equation for $\alpha$:
$$
\tan\alpha={m-n\over m+n}\cot u=
\sqrt{a^2-b^2\over a^2\sin^2u+b^2\cos^2u}\cos u,
$$
and once $\alpha$ is known we easily obtain 
$$
h=m\cos(u+\alpha)
\quad\hbox{and}\quad
k=m\sin(u+\alpha)-a.
$$
A: (corrected with angle $\alpha$ from vertical) 
Let me show an alternative - vectorial - approach.

With reference to the above sketch, and redenominating for better readability your aperture angle $u$ as $\beta$,
 the equation of the cone is given by:
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  {\bf P} = {\bf V} + \lambda {\bf v}\quad \left| {\;0 < \lambda } \right. \hfill \cr 
  {\bf v} \cdot {\bf u} =  - \cos \beta  \hfill \cr 
  \left| {\bf v} \right| = 1 \hfill \cr}  \right.
 }$$
that is:
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  {\bf v} = \left( {{\bf P} - {\bf V}} \right)/\lambda \quad \left| {\;0 < \lambda } \right. \hfill \cr 
  \left( {{\bf P} - {\bf V}} \right) \cdot {\bf u}/\lambda  =  - \cos \beta  \hfill \cr 
  \left| {\left( {{\bf P} - {\bf V}} \right)} \right| = \lambda  \hfill \cr}  \right.
 }$$
The last two equations translate into:
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  \left( {x - k} \right)\sin \alpha  + \left( {z - h} \right)\cos \alpha  =  - \lambda \cos \beta  \hfill \cr 
  \left( {x - k} \right)^{\,2}  + y^{\,2}  + \left( {z - h} \right)^{\,2}  = \lambda ^{\,2}  \hfill \cr}  \right.
 }$$
and intersecting the cone with the plane $z=0$
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  \left( {x - k} \right)\sin \alpha  - h\cos \alpha  =  - \lambda \cos \beta  \hfill \cr 
  \left( {x - k} \right)^{\,2}  + y^{\,2}  + h^{\,2}  = \lambda ^{\,2}  \hfill \cr}  \right.
 }$$
and eliminating $\lambda$ we are left with
$$ \bbox[lightyellow] {  
\cos ^{\,2} \beta \left( {x - k} \right)^{\,2}  + \cos ^{\,2} \beta y^{\,2}  + \cos ^{\,2} \beta h^{\,2}  = \left( {h\cos \alpha  - \left( {x - k} \right)\sin \alpha } \right)^{\,2} 
 }$$
which after some simple manipulations becomes:
$$ \bbox[lightyellow] {  
\left( {\cos ^{\,2} \beta  - \sin ^{\,2} \alpha } \right)x^{\,2}  + \cos ^{\,2} \beta y^{\,2}  + 2\left( {k\left( {\sin ^{\,2} \alpha  - \cos ^{\,2} \beta } \right) + 
h\sin \alpha \cos \alpha } \right)\,x = h^{\,2} \left( {\cos ^{\,2} \alpha  - \cos ^{\,2} \beta } \right) + k^{\,2} \left( {\sin ^{\,2} \alpha  -
 \cos ^{\,2} \beta } \right) + 2h\,k\sin \alpha \cos \alpha 
 }$$
For getting the required canonical equation, first of all the term in $x$ must be null
$$ \bbox[lightyellow] {  
k\left( {\sin ^{\,2} \alpha  - \cos ^{\,2} \beta } \right) =  - h\sin \alpha \cos \alpha 
 }$$
which replaced in the previous equation gives
$$ \bbox[lightyellow] {  
\left( {\cos ^{\,2} \beta  - \sin ^{\,2} \alpha } \right)x^{\,2}  + \cos ^{\,2} \beta y^{\,2}  = h^{\,2} \left( {\cos ^{\,2} \alpha  - \cos ^{\,2} \beta } \right) - \,k^{\,2} \left( {\sin ^{\,2} \alpha  - \cos ^{\,2} \beta } \right)
 }$$
Finally, to obtain the required canonical form, we are led to solve the following system of three equations in the three unknown $\alpha, k, h$
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  k\left( {\sin ^{\,2} \alpha  - \cos ^{\,2} \beta } \right) =  - h\sin \alpha \cos \alpha  \hfill \cr 
  h^{\,2} \left( {\cos ^{\,2} \alpha  - \cos ^{\,2} \beta } \right) - \,k^{\,2} \left( {\sin ^{\,2} \alpha  - \cos ^{\,2} \beta } \right) = a^{\,2} \left( {\cos ^{\,2} \beta  - \sin ^{\,2} \alpha } \right) \hfill \cr 
  h^{\,2} \left( {\cos ^{\,2} \alpha  - \cos ^{\,2} \beta } \right) - \,k^{\,2} \left( {\sin ^{\,2} \alpha  - \cos ^{\,2} \beta } \right) = b^{\,2} \cos ^{\,2} \beta  \hfill \cr}  \right.
 }$$
which simplifies to the following, that provides the final solution for $(\alpha,\, k,\, h)$ in terms of $(\beta = u,\, a, \,b)$:
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  \sin ^{\,2} \alpha  = {{\left( {a^{\,2}  - b^{\,2} } \right)} \over {a^{\,2} }}\cos ^{\,2} \beta \quad \left| \matrix{
  \;b < a \hfill \cr 
  \;\sin \alpha  < \cos \beta \quad  \Rightarrow \quad \alpha  < \pi /2 - \beta  \hfill \cr}  \right. \hfill \cr 
  k^{\,2}  = {{\left( {a^{\,2}  - b^{\,2} } \right)} \over {a^{\,2} }}{{\left( {a^{\,2} \sin ^{\,2} \beta  + b^{\,2} \cos ^{\,2} \beta } \right)} \over {\sin ^{\,2} \beta }} \hfill \cr 
  h^{\,2}  = {{b^{\,4} \cos ^{\,2} \beta } \over {a^{\,2} \sin ^{\,2} \beta }} \hfill \cr}  \right.
 }$$
example
with $a=3, \; b=2, \; \beta = u = 30^\circ$
the last identities give:
$$\sin\alpha= \sqrt{15}/6 \quad \Rightarrow \quad \alpha \approx 41.2 ^\circ$$
$$ k = \sqrt{105}/3 \approx 3.416$$
$$ h =4 \sqrt{3}/3 \approx 2.309$$
A: [For clarity, I’ve changed many of the variable names from those in the original problem. Using both $a$ and $\alpha$, and to a certain extent, $b$ and $\beta$ in the same equations is just asking for misunderstandings.]
Since I’m feeling lazy today, I’ll just use existing results instead of deriving everything from scratch. We know that when a cone is cut by a plane that does not pass through the apex, the eccentricity of the resulting conic is given by $$e={\sin\beta\over\sin\alpha},$$ with the two angles defined relative to a perpendicular to the cone’s axis as illustrated below.

(Original illustration by Ag2gaeh - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=44997120)  
In terms of our aperture half-angle $\phi$ and inclination from the vertical $\theta$, this becomes $$\sin\theta=e\cos\phi=\sqrt{1-{b^2\over a^2}}\cos\phi.\tag{*}$$ So, the tilt angle is completely determined by the aperture angle and the desired eccentricity.  
Placing the cone’s apex is thus reduced to the simple two-dimenensional problem of finding the intersection of the lines $$\begin{align} z&=(x-a)\cot(\theta-\phi) \\ z&=(x+a)\cot(\theta+\phi).\end{align}$$ Solving this system gives $$k = a{\cot(\theta-\phi)+\cot(\theta+\phi)\over\cot(\theta-\phi)-\cot(\theta+\phi)} = a{\sin2\theta\over\sin2\phi} \\ h = 2a{\cot(\theta-\phi)\cot(\theta+\phi)\over\cot(\theta-\phi)-\cot(\theta+\phi)} = a\left({\cos2\theta\over\sin2\phi}+\cot2\phi\right).$$ This can of course be expanded in terms of $a$, $b$ and $\phi$ by using equation (*).
A: It is a (somewhat) well-known fact (see this answer; Where is the cone?; Cone at MathWorld, search for "locus"; Conic Sections, section "Ellipse Seen as Circle")
that for a given ellipse, the cone from an apex $P$ to that ellipse will be circular if and only if $P$ is on a hyperbola whose foci are the vertices of the ellipse (the ends of the major axis) and whose vertices are the foci of the ellipse. (The hyperbola lies in a plane perpendicular to the plane of the ellipse, of course.)
For an ellipse given by the equation
$$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$ 
where $a > b,$ the hyperbola in the $x,z$ plane
on which the apex of the circular cone lies is simply
$$\frac{x^2}{a^2 - b^2} - \frac {z^2}{b^2}=1. \tag1$$
In order for the semiaperture angle to be $u,$ the apex of the cone must be placed on a circle whose center is at $(0,0,a \cot(2u))$ and whose radius is $a\csc(2u).$
(This is the red circle in your improved implementation on Desmos.)
That is, the apex is on the circle
$$ x^2 + (z - a\cot(2u))^2 = a^2 \csc^2 (2u) \tag2$$
in the $x,z$ plane.
To find the location of the cone's apex, solve Equations $(1)$ and $(2)$ simultaneously.

The following is an alternative method based on a geometric construction.
The figure lies entirely in the $x,z$ plane, which contains the major axis of the ellipse (denoted here as the segment $AB$) and the two foci (labeled $F$ and $F'$).

The objective is to find the apex and axis of a cone with semi-aperture angle $u$ such that the intersection of the cone with the $x,y$ plane is the given ellipse.
First, we construct an isosceles triangle $\triangle ABC$ with base
$AB$ and base angle $u.$
Construct a circle about $C$ with radius $AC.$
Let $D$ be the intersection of this circle with the line through the focus $F$ perpendicular to the major axis $AB,$ choosing the intersection nearer to $F.$ Construct the circle around $D$ with radius $DF.$
Construct the lines $AE$ and $BG$ tangent to that same circle at $E$ and $G.$
Let $H$ bet the intersection of the lines $AE$ and $BG.$
Then $H$ is the apex of the suitable cone and $HD$ is the axis.
To see this, observe that $\angle ACB = \pi - 2u,$
but due to the inscribed angles 
$\angle BAD = \beta$ and $\angle ABD = \gamma,$
we also have $\angle ACB = 2\beta + 2\gamma.$
Therefore $2u + 2\beta + 2\gamma = \pi,$
and since two angles of the triangle $\triangle ABH$ are $2\beta$ and $2\gamma,$ the third angle is $2u.$
In three dimensions, the sphere with radius $DF$ centered at $D$ is a Dandelin sphere for the intersection of the $x,y$ plane and the cone with apex $H,$ axis $HD,$ and semi-aperture angle $u.$

The geometric construction gives a quick and easy calculation of the tilt angle $\alpha.$
First, notice that $\angle ACB = \pi - 2u$ while $\angle AHB = 2u.$
Therefore the quadrilateral $ACBH$ is cyclic; in fact, it is inscribed in the red circle in the figure in the 27 Jul 2017 addendum to the question.
The axis of the cone is the bisector of $\angle AHB.$
It passes through $D$ and also (by virtue of the equal arc lengths on the red circle intercepted by inscribed angles $\angle AHC$ and $\angle BHC$) through $C.$
Therefore $CD$ lies along the axis of the cone.
But the distance from $D$ to the $z$-axis is the semi-focal distance,
$\sqrt{a^2 - b^2},$
while the distance from $C$ (which is on the $z$ axis) to $D$ is
$a \sec u.$
The tilt angle $\alpha,$ which is the angle of $CD$ from the $z$-axis, therefore satisfies
$$ \sin \alpha = \frac{\sqrt{a^2 - b^2}}{a \sec u}
= \frac{\sqrt{a^2 - b^2}}{a} \cos u,
$$
which agrees with the other answers.
Having found $\alpha,$ we can find the coordinates $(k, 0,h)$ of the cone's apex $H$ by traversing an angle $2\alpha$ from the $z$-axis around the red circle (clockwise in the figure).
The center of the circle is at $(0,0, a\cot(2u))$ and its radius is
$a/\sin(2u),$ so 
$$ k = a\frac{\sin(2\alpha)}{\sin(2u)}$$
and
$$ h = a\frac{\cos(2\alpha)}{\sin(2u)} + a\cot(2u)
= a \frac{\cos(2\alpha) + \cos(2u)}{\sin(2u)}.$$
These formulas are related to the ones in your second Desmos implementation by the relationship $\beta = \frac\pi2 - u.$
