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Question: Let $ABCD$ be a quadrilateral on the hyperbola $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Let the slopes of the sides be $m_1,m_2,m_3$ and $m_4$, then does there exist a relation between $m_1,m_2,m_3$ and $m_4$ independant of the points chosen?

My Approach: Let the eccentric angles of $(A,B,C,D)$ be $(\alpha,\beta,\gamma,\delta)$. Then:- \begin{align} m_1 &= \frac{b}{a} \frac{\cos(\frac{\alpha - \beta}{2})}{\sin(\frac{\alpha + \beta}{2})} \\ m_2 &= \frac{b}{a} \frac{\cos(\frac{\beta - \gamma}{2})}{\sin(\frac{\beta + \gamma}{2})} \\ m_3 &= \frac{b}{a} \frac{\cos(\frac{\gamma - \delta}{2})}{\sin(\frac{\gamma + \delta}{2})} \\ m_4 &= \frac{b}{a} \frac{\cos(\frac{\alpha - \delta}{2})}{\sin(\frac{\alpha + \delta}{2})} \\ \end{align} How to eliminate $\alpha,\beta,\gamma,\delta$ from these equations?

[Note: The original question was to find if such a relation holds for any conic section - I was able to show it holds for a parabola and an ellipse, but for hyperbola the calculations are very difficult.]

[Note after seeing comments: For the case of a parabola and ellipse, it is possible to extend it the property for all n-gons, where n is even. I'm not sure about the hyperbola case, as I can't seem to derive the case for a quadrilateral itself.]

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  • Since it's lack of the contexts for the cases of ellipse and parabola, I'd like to start from a circle:

    $$(x,y)=r(\cos \theta,\sin \theta)$$

    Equation of chord $AB$

    $$x\cos \frac{\alpha+\beta}{2}+y\sin \frac{\alpha+\beta}{2} =r\cos \frac{\alpha-\beta}{2}$$

    $$m_1=-\cot \frac{\alpha+\beta}{2}$$

    From properties of cyclic quadrilaterals:

    $$180^{\circ}=A+C=B+D$$

    $$0=\tan B+\tan D$$

    $$0=\frac{m_1-m_2}{1+m_1 m_2}+\frac{m_3-m_4}{1+m_3 m_4}$$

    which is equivalent to

    $$0=\tan \frac{\alpha-\gamma}{2}+\tan \frac{\gamma-\alpha}{2}$$

    $$m_1-m_2+m_3-m_4=m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3$$

  • For an ellipse $(x,y)=(a\cos \theta,b\sin \theta)$

    $$m_1=-\frac{b}{a}\cot \frac{\alpha+\beta}{2}$$

    Transform every $m_i$ back to the auxiliary circle,

    $$\frac{a}{b}(m_1-m_2+m_3-m_4)= \frac{a^3}{b^3}(m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3)$$

    $$\frac{m_1-m_2+m_3-m_4}{m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3} =\frac{a^2}{b^2}$$

    enter image description here

  • For a hyperbola $(x,y)=(a\cosh t,b\sinh t)=(a\cos it,-bi\sin it)$

    $$m_1=\frac{b}{a}\coth \frac{t+u}{2} =\frac{bi}{a} \cot \frac{i(t+u)}{2}$$

    $\fbox{Note that $t, u, v, w$ differ from the eccentricity angles $\alpha, \beta, \gamma, \delta$ in $(a\sec , b\tan )$.}$

    Now replace $b$ by $bi$, we have

    $$\frac{m_1-m_2+m_3-m_4}{m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3} =\frac{a^2}{(bi)^2}$$

    $$\frac{m_1-m_2+m_3-m_4}{m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3} =-\frac{a^2}{b^2}$$

    enter image description here

  • For a parabola of the form $x^2=4py$, take $b\to \infty$

    $$m_1-m_2+m_3-m_4=0$$

    enter image description here

  • For a parabola of the form $y^2=4qx$, take $a\to \infty$.

    $$m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3=0$$

    enter image description here

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  • $\begingroup$ Could you show the calculation for the hyperbola case? I should have mentioned it was possible to extend the result for any n-gon, where n is even for the case of a parabola and ellipse. I thought that might be the case for a hyperbola too - I can't say for sure, I can only make a guess - hence I'd like to see the calculation for a hyperbola for a quadrilateral, It might give some hint for the general n-gon case (n is even), as it did for the ellipse and parabola cases. $\endgroup$ – Kaind Sep 4 '18 at 10:58
  • $\begingroup$ I don't think it'll be the case for other $n$-gons. My considerations are based on cyclic quadrilateral. If you've more contexts, please append to your question posted. $\endgroup$ – Ng Chung Tak Sep 4 '18 at 11:01
  • $\begingroup$ Thank you, just a minor clarification required: So replacing $b$ by $ib$ in the formula for an ellipse gives the formula for a hyperbola. In this regard, if we replace 'b' by 'ib' in the slope of an ellipse we should get the slope of a hyperbola i.e $ -i \cot(\frac{\alpha + i\beta}{2}) = \frac{\cos( \frac{\alpha - \beta}{2})} {\sin( \frac{\alpha + \beta}{2})}$. Could you show how I could verify this? $\endgroup$ – Kaind Sep 4 '18 at 15:58
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    $\begingroup$ My transformation is not to ordinary auxiliary circle such that $(x,y)=(a\sec \theta,b\tan \theta)$ but in hyperbolic space $(x,y)=(a\cosh u,b\sinh u)$. Thinking of them in pure algebraic aspect. $\endgroup$ – Ng Chung Tak Sep 4 '18 at 17:36
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    $\begingroup$ Note that $\theta \ne u$. See the comparison of $(\sec, \tan)$ with $(\cosh, \sinh)$ in another answer here. $\endgroup$ – Ng Chung Tak Sep 4 '18 at 17:49
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Moving the conic and quadrilateral to the left or right won't affect the calculated slopes or the relation among them. So, let's move the conic's focus to the origin (with the closest vertex on the positive $x$-axis). Then, we can tackle all conics at once using the polar form of the equation: $$r = \frac{p}{1+\epsilon\cos\theta} \tag{1}$$ Here, $\epsilon$ is the eccentricity, and $p$ is the semi-latus rectum. Note: Slopes are coordinate-axis-dependent values, so the following analysis is specific to conics major axes parallel to the $x$-axis.

Let points $A$, $B$, $C$, $D$ be determined by $\theta$-angles $2\alpha$, $2\beta$, $2\gamma$, $2\delta$, so that we have $$A = \frac{p}{1+ \epsilon \cos 2\alpha}(\cos 2\alpha, \sin 2\alpha), \qquad\text{etc} \tag{2}$$ Then, defining $t_\theta := \tan\theta$, we can calculate various slopes: $$\begin{align} a &:= \text{slope} AB \;= \frac{\epsilon \cos(\alpha - \beta) - \cos(\alpha + \beta)}{\sin(\alpha + \beta)} = \frac{t_\alpha t_\beta(\epsilon+1)+(\epsilon-1)}{t_\alpha+t_\beta} \tag{3a}\\[4pt] b &:= \text{slope} BC \;= \phantom{\frac{\epsilon \cos(\alpha - \beta) - \cos(\alpha + \beta)}{\sin(\alpha + \beta)}} = \frac{t_\beta t_\gamma (\epsilon+1)+(\epsilon-1)}{t_\beta+t_\gamma} \tag{3b}\\[4pt] c &:= \text{slope} CD \;= \phantom{\frac{\epsilon \cos(\alpha - \beta) - \cos(\alpha + \beta)}{\sin(\alpha + \beta)}} = \frac{t_\gamma t_\delta (\epsilon+1)+(\epsilon-1)}{t_\gamma+t_\delta} \tag{3c}\\[4pt] d &:= \text{slope} DA \;= \phantom{\frac{\epsilon \cos(\alpha-\beta) - \cos(\alpha+\beta)}{\sin(\alpha + \beta)}} = \frac{t_\delta t_\alpha (\epsilon+1)+(\epsilon-1)}{t_\delta+t_\alpha} \tag{3d} \end{align}$$ One might expect that, with four equations in four unknowns, we can not eliminate all $t_i$, but it turns out that we can. (This magic seems to happen only for $2n$-gons.) Interestingly, the latus rectum factor vanishes.

$$\left(\epsilon^2-1\right)\left(\;a - b + c - d\;\right) +\left(\; b c d - a c d + a b d - a b c \;\right) = 0 \tag{4}$$ which (for non-zero $m_i$) we can write as

$$\left(\epsilon^2-1\right)\left(\;a - b + c - d\;\right) + abcd \left(\;\frac{1}{a}-\frac{1}{b}+\frac{1}{c}-\frac{1}{d} \;\right) = 0 \tag{$\star$}$$


OP mentions matter-of-factly that "it is possible to extend it the property for all $n$-gons, where $n$ is even". Here are my results for $n=6$ and $n=8$.

For hexagons, the corresponding relation is (barring typos)

$$\begin{align} 0 &= \left(\epsilon^2-1\right)^2\;\left(\;a - b + c - d + e - f\;\right) \\[4pt] &- \left(\epsilon^2-1\right) \left(\; \begin{array}{c} \phantom{-} a b c - a b d + a c d - b c d + a b e \\ - a c e + b c e + a d e - b d e + c d e \\ - a b f + a c f - b c f - a d f + b d f \\ - c d f + a e f - b e f + c e f - d e f \end{array}\;\right) \\[4pt] &- abcdef \left(\; \frac{1}{a}-\frac{1}{b}+\frac{1}{c}-\frac{1}{d}+\frac{1}{e}-\frac{1}{f} \right) \end{align} \tag{$\star\star$}$$

where a clever re-writing of the middle term currently eludes me. I'm curious about how OP's formulation of the extended property handles that term.

For octagons (again, barring typos) ...

$$\begin{align} 0 &= \left(\epsilon^2-1\right)^3 \left(\;a - b + c - d + e - f + g - h\;\right) \\[4pt] &- \left(\epsilon^2-1\right)^2\left(\; \begin{array}{c} \phantom{-} a b c - a b d + a c d - b c d + a b e - a c e + b c e + a d e \\ - b d e + c d e - a b f + a c f - b c f - a d f + b d f - c d f \\ + a e f - b e f + c e f - d e f + a b g - a c g + b c g + a d g \\ - b d g + c d g - a e g + b e g - c e g + d e g + a f g - b f g \\ + c f g - d f g + e f g - a b h + a c h - b c h - a d h + b d h \\ - c d h + a e h - b e h + c e h - d e h - a f h + b f h - c f h \\ + d f h - e f h + a g h - b g h + c g h - d g h + e g h - f g h \end{array}\;\right) \\[4pt] &+\left(\epsilon^2-1\right) \left(\; \begin{array}{c} \phantom{-} a b c d e - a b c d f + a b c e f - a b d e f + a c d e f - b c d e f + a b c d g - a b c e g \\ + a b d e g - a c d e g + b c d e g + a b c f g - a b d f g + a c d f g - b c d f g + a b e f g \\ - a c e f g + b c e f g + a d e f g - b d e f g + c d e f g - a b c d h + a b c e h - a b d e h \\ + a c d e h - b c d e h - a b c f h + a b d f h - a c d f h + b c d f h - a b e f h + a c e f h \\ - b c e f h - a d e f h + b d e f h - c d e f h + a b c g h - a b d g h + a c d g h - b c d g h \\ + a b e g h - a c e g h + b c e g h + a d e g h - b d e g h + c d e g h - a b f g h + a c f g h \\ - b c f g h - a d f g h + b d f g h - c d f g h + a e f g h - b e f g h + c e f g h - d e f g h \end{array}\;\right) \\[4pt] &+ a b c d e f g h \left(\; \frac{1}{a}-\frac{1}{b}+\frac{1}{c}-\frac{1}{d}+\frac{1}{e}-\frac{1}{f}+\frac{1}{g}-\frac{1}{h}\;\right) \end{align}\tag{$\star\star\star$}$$

Continuing the pattern is left as an exercise to the reader. (In the case of parabolas ($\epsilon=1$), it's pretty easy!)

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    $\begingroup$ The slope should read $$a=-\frac{\cos (\alpha+\beta)+e\cos (\alpha-\beta)}{\sin (\alpha+\beta)}$$ and that'll be consistent with my answer. $\endgroup$ – Ng Chung Tak Sep 4 '18 at 15:26
  • $\begingroup$ Please refer to the WA output $\endgroup$ – Ng Chung Tak Sep 4 '18 at 15:31
  • $\begingroup$ Yes, I think you wrote the slope wrong, because I know that for a quadrilateral on the parabola $y^2 = 4ax$ ($a \neq 0$), the relation is $\frac{1}{m_1} + \frac{1}{m_3} = \frac{1}{m_2} + \frac{1}{m_4}$. $\endgroup$ – Kaind Sep 4 '18 at 16:02
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    $\begingroup$ Yep. The problem with making a typo in a Mathematica script is that it propagates all the way through. :) Thanks for pointing out the embarrassing error. $\endgroup$ – Blue Sep 4 '18 at 16:15
  • $\begingroup$ For a 2n-gon, the relation for a ellipse is: $\frac{\frac{b}{a} S_1 - \frac{b^3}{a^3} S_3 + ... }{1 - \frac{b^2}{a^2} S_2 + ...} = \frac{\frac{b}{a} s_1 - \frac{b^3}{a^3} s_3 + ... }{1 - \frac{b^2}{a^2} s_2 + ...}$; where S_n is the sum of (products of (inverse of (slopes of alternate edges))) taken n times and s_n is the sum of (products of (inverse of (slopes of other set of alternate edges))) taken n times. i.e. for a hexagon, $S_1 = \frac{1}{m_1} + \frac{1}{m_3} + \frac{1}{m_5}$, $s_1 = \frac{1}{m_2} + \frac{1}{m_4} + \frac{1}{m_6}$, ... $\endgroup$ – Kaind Sep 4 '18 at 18:12
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I´m limiting myself to a quadrilateral, but this need not to be a convex one. It can be a convex quadrilateral, a concave quadrilateral and even a crossed quadrilateral. Note also that my solution doesn't employ any trigonometry.

Let $$L_1=m_1x-y+d_1=0$$ $$L_2=m_2x-y+d_2=0$$ $$L_3=m_3x-y+d_3=0$$ $$L_4=m_4x-y+d_4=0$$ be, in this order, the equations of sides AB, BC, CD, DA of a quadrilateral ABCD (it doesn't matter if convex, concave or crossed).

Then let's consider the following equation, in which k is any real number $$ kL_1L_3+L_2L_4=0$$ This is a second degree equation, therefore an equation of a conic. And an equation of a conic which circumscribes the quadrilateral ABCD, because the coordinates of every vertex satisfy both members. So the equation of this circumscribing conic takes this form: $$k(m_1x-y+d_1)(m_3x-y+d_3)+(m_2x-y+d_2)(m_4x-y+d_4)=0$$ $$(km_1m_3+m_2m_4)x^2+2(-1/2)(k(m_1+m_3)+m_2+m_4)xy+(k+1)y^2+...=0$$

Comparing it with the general second degree equation of a conic: $$A_{11}x^2+2A_{12}xy+A_{22}y^2+2A_{13}x+2A_{23}y+ A_{33}=0$$

We get these equations:

$$\begin {cases} m_1m_3k+m_2m_4=A_{11} \\ (-1/2)((m_1+m_3)k+m_2+m_4)=A_{12} \\ k+1=A_{22} \end {cases} $$

If $A_{12}=0$, $$k=-{m_2+m_4\over m_1+m_3}$$

Therefore $$A_{22}=(m_1+m_3-m_2-m_4)/(m_1+m_3)$$ $$A_{11}=\begin{vmatrix}m_2m_4 & m_1m_3 \\m_2+m_4 & m_1+m_3\\\end{vmatrix}/(m_1+m_3)$$

Thus $${A_{22}\over A_{11}} ={m_1+m_3-m_2-m_4\over \begin{vmatrix}m_2m_4 & m_1m_3 \\m_2+m_4 & m_1+m_3\\\end{vmatrix}}$$

If the conic is a circle, $A_{11}=A_{22}$ and $A_{12}=0$, and then

$${m_1+m_3-m_2-m_4\over \begin{vmatrix}m_2m_4 & m_1m_3 \\m_2+m_4 & m_1+m_3\\\end{vmatrix}}=1$$

If the conic is an ellipse with equation ${x^2\over a^2}+{y^2\over b^2}=1$, then ${A_{22}\over A_{11}} ={a^2\over b^2}$ and we get: $${m_1+m_3-m_2-m_4\over \begin{vmatrix}m_2m_4 & m_1m_3 \\m_2+m_4 & m_1+m_3\\\end{vmatrix}}={a^2\over b^2}$$

If the conic is a hyperbola with equation ${x^2\over a^2}-{y^2\over b^2}=1$, then ${A_{22}\over A_{11}} =-{a^2\over b^2}$ and we get: $${m_1+m_3-m_2-m_4\over \begin{vmatrix}m_2m_4 & m_1m_3 \\m_2+m_4 & m_1+m_3\\\end{vmatrix}}=-{a^2\over b^2}$$

But beware! The relations found above hold because xy coefficient is zero. If this coefficient is not equal to zero, we cannot state that these relations necessarily will keep valid.

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