Relationship between conic section and angle of incidence

Question

I am attempting to derive an equation that relates the area of an ellipse to its oblique cone angle alpha.

My knowns are the height and semi-major/semi-minor axis. My unknown is the angle alpha. Where do I start? Thank you.

Answer 1

I decided to add another answer because I think it is simpler than those already given.

Let $A$ and $B$ be the intersections of the ellipse with line $MO'$, with $A$ between $M$ and $O'$, and set: $p_1=AO'$, $p_2=BO'$. As explained here from the sine rule we get then:

$$ p_1={h\sin\alpha\over\sin\theta\sin(\theta+\alpha)}, \quad p_2={h\sin\alpha\over\sin\theta\sin(\theta-\alpha)}. $$ Semi-major axis $a$ can now be found as: $$ a={p_1+p_2\over2}={h\sin\alpha\cos\alpha\over \sin^2\theta\cos^2\alpha-\cos^2\theta\sin^2\alpha} $$ (where I used: $\sin(\theta\pm\alpha)= \sin\theta\cos\alpha\pm\cos\theta\sin\alpha$).

The center of the ellipse (midpoint of $AB$) has a distance $x=(p_2-p_1)/2$ from $O'$ and the perpendicular to $AB$ from $O'$ (in the plane of the ellipse) meets the ellipse at a distance $y={h\tan\alpha\over\sin\theta}$ (which is the radius of the base of the cone through $O'$). Inserting these into $x^2/a^2+y^2/b^2=1$ one gets the semi-minor axis: $$ b^2={h^2\sin^2\alpha\over \sin^2\theta\cos^2\alpha-\cos^2\theta\sin^2\alpha}. $$ We have then: $$ \sin^2\theta\cos^2\alpha-\cos^2\theta\sin^2\alpha= {h\sin\alpha\cos\alpha\over a}={h^2\sin^2\alpha\over b^2} $$ whence: $$ \tan\alpha={b^2\over ah}. $$

Answer 2

Assuming the light cone is circular

• Focal ellipse $$E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \cap z=0$$ where $a>b>0. \\$

• Focal hyperbola $$H: \frac{x^2}{c^2}-\frac{z^2}{b^2}=1 \cap y=0 $$ where $c=\sqrt{a^2-b^2}. \\$

• Radiant point

$$(\pm c\cosh v,0,b\sinh v) \in H$$

• Extremities of the major axis

$$(\pm a,0,0) \in E$$

• From the picture

\begin{align} h &= b\sinh v \\[5pt] 2\alpha &= \tan^{-1} \frac{c\cosh v+a}{h}- \tan^{-1} \frac{c\cosh v-a}{h} \\[5pt] &= \tan^{-1} \frac{c\cosh v+a}{b\sinh v}- \tan^{-1} \frac{c\cosh v-a}{b\sinh v} \\[5pt] \tan 2\alpha &= \frac{\dfrac{2a}{b\sinh v}}{1+\dfrac{c^2\cosh^2 v-a^2}{b^2\sinh^2 v}} \\[5pt] &= \frac{2ab\sinh v}{b^2\sinh^2 v+c^2\cosh^2 v-a^2} \\[5pt] &= \frac{2ab\sinh v}{a^2\sinh^2 v-b^2} \\[5pt] &= \frac{2ab^2h}{a^2 h^2-b^4} \\[5pt] &= \frac{2\left( \dfrac{b^2}{ah} \right)} {1-\left( \dfrac{b^2}{ah} \right)^2} \\[5pt] \tan \alpha &= \frac{b^2}{ah} \end{align}

• If the projection of the radiant point is within the major axis that is inside the ellipse, then

$$\frac{b^2}{a}>h>0 \iff \tan 2\alpha <0$$

• Please refer to my older posts here and here.

Answer 3

The reverse problem is easier to formulate first. I have placed an coordinate system at point M to be able to take measurements.

Given a cone with included half angle $\alpha$ offset from the x-z plane by a height $h$ and rotated by the angle $\varphi$, find the ellipse parameters described by the conic section on the x-z plane. Find semi-major axis $a$, semi-minor axis $b$, and center $c$ as shown above.

A non-rotated cone with apex at $y=h$ has the equation $x^2+z^2 = \tan^2 (\alpha) (y-h)^2$ describing all the points $(x,y,z)$ that belong on the surface of the cone.

A rotated code with rotation $\varphi$ about the z-axis has the equation

$$ \left( x \cos\varphi + (y-h) \sin \varphi \right)^2 + z^2 = \tan^2 (\alpha) \left( (y-h) \cos \varphi - x \sin\varphi \right)^2 \tag{1} $$

The equation of the ellipse described above is

$$ \left(\frac{x-c}{a}\right)^2 + \left( \frac{z}{b} \right)^2 = 1 \tag{2} $$

The set of $(x,z)$ points that obeys both equations above at $y=0$ plane is found by eliminating $z^2$ from (1) by using (2).

$$ \left( x \cos\varphi + (y-h) \sin \varphi \right)^2 + b^2 \left(1 - \left( \frac{x-c}{a} \right)^2 \right) = \tan^2 (\alpha) \left( (y-h) \cos \varphi - x \sin\varphi \right)^2 \tag{3} $$

Now I wish to find the values of $\{a,b,c\}$ that satisfy the above for all values of $x$.

First, take the 2nd derivative of (3) to group together all $x^2$ terms

$$ 2\cos^2(\varphi)-2 \left( \frac{b}{a} \right)^2 = 2 \tan^2(\alpha)\sin^2(\varphi) $$

with an intermediate solution $a = \frac{b \cos (\alpha)}{\sqrt{ \cos^2 (\alpha)-\sin^2(\varphi)}}$. Now use this in (3) and take the 1st derivative to group terms of $x$

$$ \boxed{ c = h \frac{ \sin\varphi \cos\varphi}{ \cos^2(\alpha)-\sin^2(\varphi)}} \tag{4} $$

Use the above in (3) to get

$$ \boxed{ b = h \frac{ \sin\alpha} {\sqrt{\cos^2(\alpha)-\sin^2(\varphi)}} } \tag{5} $$

and finally

$$ \boxed{ a = h \frac{ \sin \alpha \cos \alpha }{ \cos^2 (\alpha) - \sin^2 (\varphi) } } \tag{5} $$

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To reverse the process you can work out the cone angle from $a$ and $c$ and $\varphi$

$$ \alpha = \tfrac{1}{2} \sin^{-1} \left( \frac{a \sin(2 \varphi)}{c} \right) $$

Answer 4

Let the ellipse lie in the $xy$ plane with its center at $(c, 0, 0)$, where $c$ is unknown, and semi-major axis along the $x$ axis, of length $a$, and semi-minor axis along the $y$ axis, of length $b$. And let the vertex of the cone be at $V =(0, 0, h)$. The only known quantities are $a, b$, and $h$.

The parametric equation of the ellipse is

$E(t) = (c + a \cos t , b \sin t, 0 ) $

The rays connecting the vertex of the cone to the points on the ellipse are given by

$ P(t,s) = V + s (E(t) - V) $

So that

$ P - V = s (E - V) \hspace{26pt}(1)$

But,

$ E - V = (c + a \cos t , b \sin t, - h) = A u \hspace{26pt} (2) $

with

$ A = \begin{bmatrix} a && 0 && c \\ 0 && b && 0 \\ 0 && 0 && -h \end{bmatrix} \hspace{26pt}(3)$

and $ u = [\cos t , \sin t , 1 ]^T $

Vector $u$ satisfies the quadratic form

$ u^T Q_0 u = 0 \hspace{26pt}(4) $

where

$ Q_0 = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix} $

From equations (1), (2), we have

$ u = A^{-1} (E - V) = \dfrac{1}{s} A^{-1} (P - V) $

Therefore, using equation (4),

$ u^T Q_0 u = (P - V)^T A^{-T} Q_0 A^{-1} (P - V) = 0 \hspace{26pt}(5) $

From (3), we can compute

$ A^{-1} = \begin{bmatrix} \dfrac{1}{a} && 0 && \dfrac{c}{ah} \\ 0 && \dfrac{1}{b} && 0 \\ 0 && 0 && - \dfrac{1}{h} \end{bmatrix} \hspace{26pt}(6)$

Then, it follows that

$ A^{-T} Q_0 A^{-1} = Q = \begin{bmatrix} \dfrac{1}{a^2} && 0 && \dfrac{c}{a^2 h} \\ 0 && \dfrac{1}{b^2} && 0 \\ \dfrac{c}{a^2 h} && 0 && \dfrac{1}{h^2} \left (\left(\dfrac{c}{a}\right)^2 - 1\right) \end{bmatrix} \hspace{26pt}(7)$

To simplify the notation, let

$ r = \dfrac{c}{a}, A = \dfrac{1}{a^2}, B = \dfrac{1}{b^2} , H = \dfrac{1}{h^2} $

Then

$ Q = \begin{bmatrix} A && 0 && r \sqrt{AH} \\ 0 && B && 0 \\ r \sqrt{AH} && 0 && H (r^2 - 1) \end{bmatrix} \hspace{26pt}(8)$

The characteristic polynomial of Q is

$ F(\lambda) = (\lambda - B) (\lambda^2 - ( A + H (r^2 - 1) ) \lambda - A H ) \hspace{26pt}(9) $

Note that by Vieta, the product of the eigenvalues resulting from the quadratic factor is $(-AH)$ which is negative. To get a right circular cone we need two equal positive eigenvalues and one negative eigenvalue. Hence we require that the positive root of the quadratic factor of the characteristic polynomial to be equal to $B$. Hence,

$ B = \dfrac{1}{2} ( A + H (r^2 - 1) + \sqrt{ (A + H (r^2 - 1) )^2 + 4 A H } $

From which

$ \bigg( 2 B - (A + H(r^2 - 1) ) \bigg)^2 = (A + H (r^2 - 1) )^2 + 4 A H $

And this yields,

$ 4 B^2 - 4 B ( A + H(r^2 - 1) ) = 4 A H $

So that

$ B^2 - B (A + H (r^2 - 1) ) = A H \hspace{26pt}(10) $

And this equation can be solved for $r$, which gives us the distance between the center of the ellipse and the origin. (recall that $r = \dfrac{c}{a} $ ).

So now, we have

$ \lambda_1 = \lambda_2 = B $

We can now compute the third (negative) eigenvalue as follows

$ \lambda_3 = \dfrac{1}{2} ( A + H (r^2 - 1) - \sqrt{ (A + H (r^2 - 1) )^2 + 4 A H } $

From (10), we have

$ A + H (r^2 - 1) = B - \dfrac{AH}{B} $

So,

$ \lambda_3 = \dfrac{1}{2} \bigg( B - \dfrac{AH}{B} - \bigg( B + \dfrac{AH}{B} \bigg) \bigg) =- \dfrac{AH}{B} \hspace{26pt}(11) $

Therefore, our $Q$ matrix is similar to

$ Q' = \begin{bmatrix} \lambda_1 && 0 && 0 \\ 0 && \lambda_2 && 0 \\ 0 && 0 && \lambda_3 \end{bmatrix} \hspace{26pt}(12)$

Note that $\lambda_1 = \lambda_2 = B \gt 0 $ and that $ \lambda_3 = - \dfrac{AH}{B} \lt 0 $

Matrix $Q'$ represents the cone

$$ v^T Q' v = 0 $$

where the vector $v$ is given by

$$ v = ( \sqrt{- \dfrac{\lambda_3}{\lambda_1} } \cos \phi , \sqrt{- \dfrac{\lambda_3}{\lambda_1}} \sin \phi, 1 ) \tag{13} $$

Thus the semi-vertical angle $\alpha$ is given by

$$ \tan \alpha = \sqrt{ - \dfrac{ \lambda_3}{\lambda_1} }\tag{14}$$

But $$ -\dfrac{ \lambda_3}{\lambda_1} = \dfrac{A H}{B^2} = \dfrac{b^4}{a^2 h^2} \tag{15} $$

Hence,

$$ \boxed{\tan \alpha = \dfrac{b^2}{a h} } \tag{16} $$