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For a given ellipse, find the locus of all points P for which the two tangents are perpendicular.

I have a trigonometric proof that the locus is a circle, but I'd like a pure (synthetic) geometry proof.

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two tangents? Tangents to what? perpendicular to what? –  Willie Wong Apr 17 '11 at 20:08
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@Willie: The points at which the two tangents to the ellipse through that point are perpendicular to each other. –  Arturo Magidin Apr 17 '11 at 20:40
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As an additional note: what these answers show is that the orthoptic of an ellipse is a circle. The result is sometimes attributed to Monge. See this paper for instance. –  J. M. Apr 18 '11 at 0:07
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Please excuse my English that is probably not quite correct.

Here is a proof that I read somewhere (I am not the finder of it). It may seem long, but it is because of the way I explain it ; actually it is very simple. Is it "synthetic" as you wish ? You will judge yourself.

First of all, imagine a rigid plane that glides on a fixed plane, and consider, at a fixed time t, the vector field of the velocities, with respect to the fixed plane, of all the points of the moving plane. Since the moving plane is rigid, the vector field of the velocities must be of one of the following two types : (i) or it is a constant vector field, and that means that all the points of the moving plane have the same (vectorial) velocity at time t (case of an instant translation) ; (ii) or there is a point C, called the instant centre of rotation, such that the velocity at any point M is orthogonal to the vector CM (case of an instant rotation). Of course, the instant centre of rotation may move to different places during the movement. For a proof of these facts, see the appendix below.

Now draw your ellipse on a fixed plane, and draw a pair of orthogonal axes crossing at point A on some other rigid plane. Then put the second plane on the first one in such a way that the axes are tangent to the ellipse, and finally rotate this second plane on the fixed one while keeping the axes tangent to the ellipse. What we are looking for is the locus of point A during this movement.

Next fix a point P and a point Q on the axes such that, at a given fixed time t, these points coincide with the points of contact of the axes with the ellipse. My first claim is that at that very same time t, the velocities of points P and Q are tangent to the ellipse. Indeed, if they were not, at least one of these points would cross the ellipse at time t, and this would lead to a contradiction with the fact that the axes remain tangent to the ellipse. From this first claim, and since these velocities are orthogonal to each other, we may conclude that there is an instant centre of rotation C located at the intersection of the normals to the ellipse at points P and Q, or in other words, located at the fourth vertex of a rectangle with first three vertices at P, A and Q. It follows that the velocity of point A at that same time t is orthogonal to the diagonal CA of the rectangle (PAQC).

My second claim is that this diagonal CA passes through the centre O of the ellipse. Indeed, this follows from the fact that it passes through the midpoint of P and Q since (PAQC) is a rectangle, and from the well-known fact that when given two tangents (even non orthogonal) to an ellipse, then the intersection point of the two tangents, the midpoint of the two points of contact, and the centre of the ellipse, all lie on a same straight line : it is obvious by symmetry when the ellipse is a circle, and all these facts can be transported to any ellipse since they remain invariant under affine transformation. As a consequence of this second claim, we get that the velocity of point A at time t is orthogonal to the vector OA.

But this result holds for any fixed time t. Therefore, the velocity of point A always remains orthogonal to the vector OA, and it follows that the (square of the) distance OA remains constant during the movement since it has a zero derivative. And this proves that point A moves along a circle centred at O.

The end.

Appendix : the vector field of the velocities of a rigid moving plane.

Let us denote by M' the velocity of point M at some fixed time t. If A and M are two points of the rigid moving plane, by differentiating the square of the constant distance AM we get that M'-A' is orthogonal to the vector AM. Let us mention two consequences of this : (i) if there is a point C such that C'=0, then M' is orthogonal to the vector CM for all points M of the moving plane (case of an instant rotation) ; (ii) if A is such that A'$≠$0, then for all points M lying on the line through A orthogonal to A', the velocities M' are parallel to A' (ie they are linearly dependent).

Then we may conclude as follows : (i) if there are two points A and B such that A' and B' are not parallel (ie are linearly independent), the lines through A orthogonal to A' and through B orthogonal to B' cross each other at some point C where the velocity must be parallel to both A' and B', so that C'$=0$, and then we have an instant rotation as proved above ; (ii) and if (otherwise) the velocities are all parallel to some direction A'≠0, then at any point M such that AM is not orthogonal to A' the velocity has to be M'=A', and it is not difficult to see that this implies that actually M'=A' for all points M of the moving plane, that is, then we have an instant translation.

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Very nice! I find it very intuitive: (1) As you move the T-square around the ellipse, its instantaneous center of rotation must lie on the perpendiculars to the tangent points, and (2) this center of rotation C and the center O of the ellipse both lie on the line connecting the T-square's corner A with the midpoint M of the 2 tangent points (to see it for O, transform the ellipse affinely to a circle). (3) Therefore A never changes its distance from O. Beautiful! I don't know whether the idea of instantaneous motion was around in ancient Greece, but the proof definitely has a synthetic feel. –  Matt May 4 at 10:24
    
This approach also works to show that the locus, in the case of a parabola, is a straight line (the directrix), using the fact that rays from the focus are reflected to be parallel, and completing the rhombus. This same proof technique can also be used to give an alternate proof of the well-known fact that the locus of points A from which segment EF subtends a given angle is a circle: The instantaneous rotation is always around the diametrically opposite point of the circle. A rather powerful technique! –  Matt May 4 at 13:42
    
The same method can be used to show that the same question, in the case of a suitably eccentric hyperbola (instead of an ellipse), also yields a circular locus, centered at the hyperbola's center. For this proof, the affine transformation step should put A onto the major or minor axis of the hyperbola (depending on whether the tangents were on the same or different branches of the hyperbola). This is again called the director's circle. –  Matt May 4 at 13:54
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It is a well known circle, generally called the "Director circle", though Wikipedia prefers "Fermat–Apollonius circle"

You can find a projective geometric proof here from Paris Pamfilos though I prefer one based on analytical geometry such as the discussion here by Michael Raugh or the following from slides 12 and 13 from Career Launcher India using earlier results for tangents to an ellipse:

If the ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and we take a tangent to the ellipse from the point $(h,k)$ with slope $m$ then $(h^2-a^2)m^2-2hkm+(k^2-b^2) =0$.

This has two solutions $m_1$ and $m_2$ with $m_1 m_2 = \dfrac{k^2-b^2}{h^2-a^2}$, and the two tangents are perpendicular if $m_1 m_2 = -1$ so we need $h^2 +k^2 = a^2+b^2$, meaning the locus is the circle $$x^2+y^2 = a^2+b^2.$$

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I didn't realize the circle has a name! The Pamfilos page is the sort of thing I was hoping for, but I'm missing something simple on the very first step: Why is it obvious that circle OP, circle GHG'H', and circle MNM'N' are all the same circle? (M,N,G, etc. being defined as projections of the foci onto the tangents.) The steps after that are fine. –  Matt Apr 18 '11 at 0:17
    
@Matt: you might sometimes see it referred to as the "Monge circle" as well. –  J. M. Apr 18 '11 at 0:37
    
The name corresponds to a directrix for a parabola. I think the monograph referred to may be here but it is in Modern Greek and the pictures do not have circles [Adobe also suggests there is only 1 page, but there are up to 266]. –  Henry Apr 18 '11 at 0:47
    
Ah yes, page 174 (page 186 using their pdf reader) of the post-graduate thesis by Μπουνάκης, Δημήτρης in your link gives the answer. –  Matt Apr 18 '11 at 9:51
    
Specifically, it proves that the projection of a focus onto a tangent will lie on the "auxiliary circle" (the circle having the ellipse's major axis as a diameter), like this: Let E, F be the foci, O the center, T the point of tangency, and Q the projection of E onto the tangent ("projection" meaning EQ ⊥ TQ). Extend EQ and FT to meet at R. It is a property of ellipses that ∠ETQ=∠RTQ (i.e., they reflect sound/light from one focus to the other), so ▵ETQ≅▵RTQ. So QO is the midline of ▵REF, so 2 QO = RF = ET+TF = diameter of circle, making QO a radius of the circle. QED. –  Matt Apr 18 '11 at 10:48
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If all you want is a proof that the locus is a circle, we may assume that the ellipse is given by $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$

Ignoring vertical tangents for now, if a line $y=mx+k$ is tangent to the ellipse, then plugging in this value of $y$ into the equation for the ellipse gives $$\frac{x^2}{a^2} + \frac{(m^2x^2 + 2mkx + k^2)}{b^2} = 1$$ or $$(b^2 + a^2m^2)x^2 + 2a^2mkx + (a^2k^2 - a^2b^2) = 0.$$ This equation gives the two points of intersection of the line with the ellipse. If the line is tangent, then the two points must coincide, so the quadratic must have zero discriminant. That is, we need $$(2a^2mk)^2 - 4(a^2k^2 - a^2b^2)(b^2+a^2m^2) = 0$$ or equivalently, $$\begin{align*} (a^2m^2)k^2 -a^2(b^2+a^2m^2)k^2 &= -a^2b^2(b^2+a^2m^2)\\ -a^2b^2k^2&= -a^2b^2(b^2+a^2m^2)\\ k^2 &= b^2+a^2m^2\\ k &= \pm\sqrt{a^2m^2 + b^2}. \end{align*} $$ So the lines that are tangent to the ellipse are of the form $$y = mx \pm \sqrt{a^2m^2 + b^2}.$$

Since the problem is symmetric about $x$ and $y$, consider the points on the upper half plane, so that we will take the plus sign above. The tangent perpendicular to this one will therefore have equation $$y = -\frac{1}{m}x + \sqrt{\frac{a^2}{m^2} + b^2},$$ or equivalently $$my = -x + \sqrt{a^2 + m^2b^2}.$$ (We are ignoring the vertical and horizontal tangents; I'll deal with them at the end).

If a point $(r,s)$ is on both lines, then we have $$\begin{align*} s-mr &= \sqrt{a^2m^2 + b^2}\\ ms + r &= \sqrt{a^2+m^2b^2}. \end{align*}$$ Squaring both sides of both equations we get $$\begin{align*} s^2 - 2mrs + m^2r^2 &= a^2m^2 + b^2\\ m^2s^2 + 2mrs + r^2 &= a^2 + m^2b^2 \end{align*}$$ and adding both equations, we have $$\begin{align*} (1+m^2)s^2 + (1+m^2)r^2 &= (1+m^2)a^2 + (1+m^2)b^2,\\ (1+m^2)(s^2+r^2) &= (1+m^2)(a^2+b^2)\\ s^2+r^2 = a^2+b^2, \end{align*}$$ showing that $(s,r)$ lies in a circle, namely $x^2+y^2 = a^2+b^2$.

Taking the negative sign for the square root leads to the same equation.

Finally, for the vertical and horizontal tangents, these occur at $x=\pm a$; the horizontal tangents are $y=\pm b$. Their intersections occur at $(\pm a,\pm b)$, which lie on the circle given above. So the locus of such points is contained in the circle $x^2+y^2 = a^2+b^2$.

Conversely, consider a point $(r,s)$ that lies on $x^2+y^2 = a^2+b^2$. If a tangent to the ellipse $$ y = mx + \sqrt{a^2m^2 + b^2}$$ goes through $(r,s)$, then we have $$ s-mr = \sqrt{a^2m^2+b^2}.$$ Squaring both sides, we have $$s^2 - 2msr + m^2r^2 = a^2m^2 + b^2$$ or $$(a^2-r^2)m^2 +2srm + (b^2-s^2) = 0.$$ Since $r^s+s^2 = a^2+b^2$, then $a^2 - r^2 = s^2-b^2$, we we have $$(s^2-b^2)m^2 + 2srm + (b^2-s^2) = 0,$$ and if we do not have $s=\pm b$ (the horizontal/vertical tangent intersection points), then we get $$m^2 + tm - 1 = 0,\qquad\text{where } t = \frac{2sr}{s^2-b^2}.$$ So the two solutions for $m$, $m_1$ and $m_2$, satisfy $m_1m_2 = -1$, hence the two tangents are perpendicular. That is, at every point on the (upper half of the) circle, the two lines through the point that are tangent to the ellipse are perpendicular to each other.

So all such points are on the circle, and all points on the circle are such points. (The circle is called the director circle of the ellipse).

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Not a synthetic proof, but ...

It's interesting to note that four target points are easy to identify: these are the corners of the rectangle that bounds the ellipse on the left, right, top, and bottom; call this the "axis-aligned bounding rectangle". Clearly, if we rotate the ellipse about its center, then the corners of the (new, un-rotated) axis-aligned bounding rectangle of the rotated figure also correspond to four target points. We can show that the target points form a circle by showing that these axis-aligned rectangles always have semi-diagonal $\sqrt{a^2 + b^2}$, regardless of rotation.

Parameterizing the ellipse by $P(a\cos\theta,b\sin\theta)$ and rotating about the center by angle $\phi$ gives an ellipse parameterized by

$$(a \cos\theta \cos\phi - b \sin\theta \sin\phi, a \cos\theta \sin\phi + b \sin\theta \cos\phi )$$

The right side of the axis-aligned bounding rectangle is determined by the value of $\theta$ that maximizes the $x$ coordinate; the top side is determined by the value that maximizes the $y$ coordinate.

Note that both parameterized coordinates have the form $U\cos\theta + V\sin\theta$, which can be expressed as $\sqrt{U^2+V^2} \sin\left(\theta+\psi\right)$ for some $\psi$. As $\theta$ traverses all values from $0$ to $2\pi$, the coordinate formula must attain a maximum value of $\sqrt{U^2 + V^2}$ (and we needn't worry about the exact value of $\psi$).

Consequently, the rotated ellipse's maximal $x$-coordinate is $x_{\star}:=\sqrt{a^2\cos^2\phi + b^2\sin^2\phi}$ and its maximal $y$-coordinate is $y_{\star}:=\sqrt{a^2\sin^2\phi + b^2\cos^2\phi}$, so that the rectangle's semi-diagonal, $d$, satisfies

$$d^2 = x_{\star}^2 + y_{\star}^2 = \left( a^2 + b^2 \right)\left( \cos^2\phi + \sin^2\phi \right) = a^2 + b^2$$

Therefore, the corners of the axis-aligned bounding rectangles of the rotated ellipse lie on the circle of radius $\sqrt{a^2 + b^2}$, regardless of the angle of rotation, $\phi$; rotating them back through angle $\phi$ won't change that fact.

This shows that all of the points with the "perpendicular tangents" property lie on a particular circle, but it does not show that all points on that circle have the "perpendicular tangents" property. That implication is left to the reader.

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If you hadn't posted this, I'd have done it this way. :) –  J. M. Apr 17 '11 at 23:58
    
This is exactly my "trigonometric" proof. (I had optimistically resized the ellipse so that $a = cos(\gamma)$ and $b = sin(\gamma)$, making all of the formulas entirely built of trig functions, and at the last step the $a^2+b^2$ also disappears.) Very nice explanation. –  Matt Apr 18 '11 at 9:26
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