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.

enter image description here

A screenshot from a desmos simulation is shown below.

$\hspace{2cm}$[enter image description here]2


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 $b$.

enter image description here

  • $\begingroup$ It shouldn’t come as a surprise that this is similar to the martini glass problem. For both of them the crux is computing the ellipse created by the intersection of a given cone and a plane. $\endgroup$ – amd Jul 11 '17 at 18:36
  • 1
    $\begingroup$ This is related to the general question of given a conic section on a plane, find all the cones in 3D that would generate this section. Essentially find the light source position and sweep angle. I think this is a very interesting question mathematically. $\endgroup$ – ja72 May 10 at 11:29
  • $\begingroup$ @ja72 That sounds like an interesting problem. Has it been posted here before? What is the locus of the vertex of all such cones? $\endgroup$ – hypergeometric May 12 at 6:45
  • $\begingroup$ @hypergeometric - I haven't posted it. It came about when I realized that a planar parabola looks like an ellipse in perspective view. I was trying to visualize the 3D cones that generate specific conic sections. $\endgroup$ – ja72 May 12 at 22:20

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 $AVB$: $$ m^2+n^2-2mn\cos2u=4a^2. $$

enter image description here

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 $$ h=m\cos(u+\alpha) \quad\hbox{and}\quad k=m\sin(u+\alpha)-a. $$

  • $\begingroup$ I improved and corrected the formula for $\tan\alpha$. Everything should be fine now, $\endgroup$ – Aretino Jul 9 '17 at 7:35
  • $\begingroup$ Nice solution, esp the circle.(+1) $\endgroup$ – hypergeometric Jul 9 '17 at 17:06
  • $\begingroup$ Did you or others,by any chance, attempt animating variable vertex martini glass tips (spilling martini position) using Manipulate inclination sliders on Mathematica site?. Using ParametricPlot3D? I'd love to see it. That would perhaps be a wee bit more realistic than what we could see on desmos. $\endgroup$ – Narasimham May 29 at 18:02
  • $\begingroup$ @Narasimham I didn't try, but I'd rather use GeoGebra than Mathematica. $\endgroup$ – Aretino May 29 at 18:22
  • $\begingroup$ @Aretino Thanks. GeoGebra is fine. Thought would be more comfortable on Mma playing with other wine glass meridians if required. $\endgroup$ – Narasimham May 29 at 18:31

[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.

enter image description here

(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 (*).

  • $\begingroup$ Nice solution. (+1) $\endgroup$ – hypergeometric Jul 12 '17 at 15:56

(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 $(\beta = u,\, a, \,b)$: $$ \bbox[lightyellow] { \left\{ \matrix{ \sin ^{\,2} \alpha = {{\left( {a^{\,2} - b^{\,2} } \right)} \over {a^{\,2} }}\cos ^{\,2} \beta \quad \left| \matrix{ \;b < a \hfill \cr \;\sin \alpha < \cos \beta \quad \Rightarrow \quad \alpha < \pi /2 - \beta \hfill \cr} \right. \hfill \cr k^{\,2} = {{\left( {a^{\,2} - b^{\,2} } \right)} \over {a^{\,2} }}{{\left( {a^{\,2} \sin ^{\,2} \beta + b^{\,2} \cos ^{\,2} \beta } \right)} \over {\sin ^{\,2} \beta }} \hfill \cr h^{\,2} = {{b^{\,4} \cos ^{\,2} \beta } \over {a^{\,2} \sin ^{\,2} \beta }} \hfill \cr} \right. }$$


with $a=3, \; b=2, \; \beta = u = 30^\circ$
the last identities give: $$\sin\alpha= \sqrt{15}/6 \quad \Rightarrow \quad \alpha \approx 41.2 ^\circ$$ $$ k = \sqrt{105}/3 \approx 3.416$$ $$ h =4 \sqrt{3}/3 \approx 2.309$$

  • $\begingroup$ Nice solution (+1). Close to what I have, except that the $\alpha$ as defined in the question is the tilt from vertical (such that if $\alpha=0$ then the resultant spotlight is circular) whereas you have defined it as the complementary angle. Would you be able to modify your solution for easy of comparability to the question as posted and to other solutions here? $\endgroup$ – hypergeometric Jul 9 '17 at 17:08
  • $\begingroup$ Ah, sorry, then just to change $\cos \alpha$ with $\sin \alpha$., and v.v.. I'll amend my answer. $\endgroup$ – G Cab Jul 9 '17 at 18:03
  • $\begingroup$ @hypergeometric: answer amended with $\alpha$ taken from vertical. $\endgroup$ – G Cab Jul 10 '17 at 11:15

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 $F'$).

enter image description here

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 $\beta = \frac\pi2 - u.$


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