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A non-violent anarchist Mathematics teacher and Christian who has views about everything under the sun.

  • A Festival of Flipping Lights

    Presenting Problem

    (Source: The New York Times)

    I recently came across the following problem: Eight lamps, each with an on-off switch, are arranged in a circle. A lamplighter can flip (in other words, change the state of) the switches, but he is not allowed to flip just one at a time. When he switches lamps on or off, he is required to flip the switches of three consecutive lamps simultaneously. (For example, he can flip the switches of lamps G, H and A at the same time.) Prove that no matter what set of lamps was turned on at the start, the lamplighter can turn all the lamps on.

    Recognizing the Requirement

    Since we are entering the Christmas season, which is a season of lights, I wish to tackle this specific problem and then extend it to a more general situation. Now, since the final state of all the lights is the same with all being on is accomplished regardless of the initial state of the lights, it must mean that it is possible to change the state of any particular light after a finite number of moves. Then we can repeat the process as needed to change the state of every light that is currently off so that it is on. In effect this means that we are able to transform any starting pattern to any other desired pattern with a finite number of steps, each of which consists of flipping three switches.

    Initial Proof

    To begin with let us just assume that all the lights are off. This is shown in the figure below.

    Here, I am adopting the convention that a filled-in circle represents a light that is on. Hence, an empty circle is a light that is off. Now, the challenge is to use a finite number of moves to change the state of one and only one of the lights. Let us arbitrarily start at A. Hence, we have to flip the switches A, B, and C. This will give us

    Now, we can flip the switches D, E, and F to give us

    Finally, we can flip the switches G, H, and A to give

    We can see that, after three switching sets we have actually managed to alter the state of all the lights except A. If we continue this process, the next two outcomes will be:

    Note that, starting with A, after 5 switching sets, we have managed to alter the state of only H. This is because, 5 switching sets involves 15 switching elements, which is 1 less than 2 times 8. Hence, all the lights from A to G get switched twice, restoring them to their original state, while H gets switched only once, thereby flipping its state. In other words, if I desire to flip the state of one of the lights, all I need to do is start with its successor and perform 5 switching sets. Let us test this on an example.

    Verification

    Since there are 256 (i.e. 28) different permutations of on and off lights in the set of 8, I used a random number generator to first choose a number from 1 to 256 and then iterated the random number generator that many times. The first number it spat out was 55 and I then generated 55 more numbers. This gave me 158, which in binary is 10011110. Hence, the starting case will be lights B, C, D, E, and H switched on. This is shown below.

    We will represent this state as aBCDEfgH, where the lower case letters indicate that the light is switched off and the upper case letters indicate it is switched on. Hence, we have to switch the lights A, F, and G on to achieve our goal. We will start with B, which is the successor of A. Performing 5 sets of switching we obtain the following states:

    Note that, even though the goal was to change the state of only A, which we manage in step 5 as expected, step 4 actually gives us the final state we wanted. For now, we will ignore any occasions when the final state is accomplished in the middle and proceed only with the state after 5 sets are complete. Hence, we will continue with the state ABCDEfgh. Now F needs to be flipped. So we start with G and the next 5 sets will give us

    Now, we need to flip G. So starting with H the next 5 sets give us

    So, we have managed to flip light G on as well, leaving only light H off. Now, we start with A and the next 5 sets will give us

    We have, therefore, achieved the final state we wanted with all lights switched on. Note, that the initial state I chose was done at random. Indeed, our algorithm indicates that, starting with any particular light, after 5 sets of switching its predecessor will be the only one to have changed states.

    Bezout’s Identity

    Why does this happen? Is this limited to the numbers 8 (for number of lights) and 3 (for number of consecutive lights in each set)? Of course not! Rather, what we see here is the fact that

    Here, the ‘2’ indicates the number of times all except one of the lights have their states changed and the ‘5’ represents the number of sets that need to be performed to isolate one light and flip only its state. Is it possible to generalize this?

    Indeed it is! This is an example of an application of what is known as Bezout’s identity. Étienne Bézout was a French mathematician who did significant work on number theory. He demonstrated that, if a and b are two natural numbers, then there exist integers p and q such that

    In our example, a = 8 and b = 3, for which gcd(8, 3) = 1. Here we have found p = 2 and q = -5 such that

    Testing a Hypothesis

    We can readily see that the reason we were able to isolate one of the lights was precisely because 5 sets of 3 each put us 1 short of a multiple of 8. Hence, can we propose that if we choose a lights and restrict ourselves to flipping sets of b consecutive lights, where gcd(a, b) = 1, then we should be able to isolate any light by conducting q set flips, which will accomplish p changes of state to all but 1 light? Let us test this hypothesis with an example.

    Suppose now we have 11 lights and restrict ourselves to flipping 4 consecutive lights. Let us begin with the state abcdefghijk, where all the lights are off. We observe that

    Now, starting with A and performing 3 set flips, we will get

    We have indeed isolated light A. However, rather than flipping the state of only A, we have managed to flip the state of all lights except A! Can we rectify this? Of course, we can! Bezout’s identity does not say that the pair (p, q) is unique. Indeed, we can observe that

    This seems to indicate that we should be able to isolate A and change its state. When we perform 8 flip sets we will get the following.

    Once again we observe that we have indeed isolated one of the lights, in this case K. And now too we have left it unflipped while flipping all the others. Are we stuck?

    Let us try to interpret the numbers in the equation

    The ’11’ and ‘4’ are clear. There are 11 lights and we are forced to flip 4 consecutive lights. The ‘8’ indicates that we will perform 8 flip sets. But the ‘3’ indicates that, when we have isolated one of the lights, all the others would have undergone ‘3’ state changes. Hence, all the lights, except the one we are focusing on actually change their states, leaving the one we are focusing on with an unchanged state. Hence, with 11 lights and flip sets of 4 consecutive lights it is not possible to isolate 1 light and change one its state. It is possible to leave only its state unchanged.

    Indeed, if 11p + 4q = 1 has to hold for integers p and q then it is necessary for p to be odd. However, p indicates the number of state changes that all lights other than the one we are focusing on undergo. This means that all the other lights will change states and the one we are focusing on remains unchanged.

    Modifying the Hypothesis

    For us to be able to isolate 1 light for a state change, it is necessary but not sufficient that ap + bq = 1. In addition, p must be even so that all other lights return to their original states while the light we are focusing on changes its state. However, from ap + bq = 1 we obtain

    If p is even, this means that both b and q must both be odd because only then can the difference 1 – bq be even. So let’s assume that a = 12 and b = 5. The GCD of 12 and 5 is 1. Will that allow us to isolate any particular light? From the above formula, we can see that

    Hence, with a = 12 lights and b = 5 being flipped each time, we should need q = 5 sets, each, except the one we start with, being flipped p = 2 times. Let’s test this starting with abcdefghijkl, the situation where all 12 lights are off. Carrying on with flip sets of 5, we can obtain the following

    As can be seen, we have isolated light A and changed only its state without affecting the states of any of the others. This will allow us to change any pattern to any other pattern we desire.

    Stringing Along

    What we have seen is that, if the number of lights is odd, then we are not guaranteed that there is a strategy that could convert any given pattern to any other pattern. However, if the number of lights is even and the number we flip each time is coprime with the number of lights then we may be to alter any starting pattern to any other pattern. However, this is not guaranteed. Is there a way of obtaining a generalized solution to this? That is, can we determine the exact conditions for a, the number of lights, and b, the number of lights flipped each time, that will always give us the possibility of changing any starting pattern to any other pattern?

  • Identifying Linear Commonalities

    The Final Spur

    For some weeks now, we have been focusing on the conic sections. In the last post, A Mathematical Straight Couple, we introduced the pair of lines as a degenerate hyperbola. We saw that the joint equation of a pair of lines through the origin is

    This kind of equation, where the sum of the powers of the variables in each term is the same and the constant term is zero is known as a homogeneous equation. However, the general second degree equation in two variables, which is

    is not a homogeneous equation. For this equation, in the previous post, we defined

    and we showed that if ∆ = 0 the general second degree equation in two variables represents a pair of lines. In this post, I wish to look at determining the point of intersection of a pair of lines described in the form of the general second degree equation in two variables and to look at another concept known as ‘homogenization’.

    Transformation of Coordinates

    Before we can determine the point of intersection of the two lines, we need to learn about transformation of coordinates. Suppose we have a point A(x, y) according to a previously defined set of rectangular axes. Now suppose we move the origin and the axes (with translation only, no rotation) to the point (p, q). It is evident that the coordinates of all points on the plane will change. Suppose the new coordinates of A are (X, Y). This is shown in the diagram below.


    From the diagram above it is clear that X = x – p and Y = y – q. Hence, when we move the axes p units to the right and q units up, the new x and y coordinates, indicated by capital letters in our convention, are reduced by p and q respectively.

    The Point of Intersection of the Lines ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

    Suppose the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents two straight lines and that their point of intersection is (p, q). Now, if we shift the origin to the point of intersection, we should get a homogeneous equation in the new coordinates because the lines now pass through the new origin. From what we know about transformation of coordinates X = x – p ⇒ x = X + p and y = Y + q. Making these substitutions we get

    Grouping terms according to degree and variable we get

    If this is a homogeneous equation it must reduce to

    Then we must have

    Now since (p, q) is the point of intersection of the lines, the coordinates must satisfy the joint equation of the lines. Hence,

    Hence, the third equation above is automatically satisfied by the point of intersection of the two lines. Solving the equations ap + hq + g = 0 and hp + bq + f = 0 simultaneously we get

    which gives the coordinates of the point of intersection.

    Coincident and Parallel Lines

    Recall, from the previous post, that the angle between the lines

    is given by

    It is clear that, if h2 – ab = 0, then the angle between the lines is 0°, meaning that the lines are either coincident or parallel. In that case, there will be infinitely many points common to the two lines or no points in common. Hence, we can say that, if ∆ = 0, h2 – ab = 0, bg – hf = 0, and af – gh = 0, then the lines are coincident. Also, if ∆ = 0, h2 – ab = 0 and either bg – hf ≠ 0 or af – gh ≠ 0, then the lines will be parallel.

    Homogenization of a Curve with a Line

    Suppose we have a curve S = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 and a line L = lx + my – n = 0. These are shown in the figure below in red and green respectively.

    Let the line and the curve intersect at A and B as shown. It is evident that the coordinates of A and B satisfy the equations of both the curve and the line. Hence, A and B satisfy any combination of the two equations. Suppose we write the equation of the line as

    This is true for points A and B. Hence,

    holds at points A and B. Hence,

    However, this last equation is a second degree homogeneous equation in two variables and must represent two straight lines through the origin O that also pass though points A and B. Hence, this last equation represents the joint equation of the lines OA and OB.

    Now, the concept of homogenization might seem to be strange and perhaps even difficult to use. So, let us actually use it to solve a problem.

    Using Homogenization

    Let us prove that chords of the parabola y2 = 4ax which subtend a right angle at the origin pass through a fixed point. Note that what follows is a standard 4-step approach to problem solving, which I hope will help some of the readers since it works in pretty much every area of life and not just mathematics!

    Step 1: Information collection
    Given the parabola y2 = 4ax. Let the chord be lx + my – 1 = 0. We have to prove that if the chord subtends a right angle at the origin then it must pass through a fixed point.
     
    Step 2: Conceptual plan
    If we homogenize the equation of the curve S = y2 = 4ax with the equation of a line L = lx + my – 1 = 0 then we obtain the joint equation of the lines joining the origin to the points of intersection of the curve S and the line L. Further, the lines ax2 + 2hxy + by2 = 0 are at right angles to each other if a + b = 0.

    Step 3: Execution
    Homogenizing with the equation of the parabola we get

    This represents a pair of lines through the origin. If the lines are perpendicular to each other

    Hence, the line is

    where m is a parameter. This represents a family of lines all of which pass through the point of intersection of the lines x – 4a = 0 and y = 0. But the point of intersection of these two lines is (4a, 0). That is, the chord passes through the point (4a, 0)

    Step 4: Verification
    The working is correct and we have proved what had to be proved. And note that we did not even need to know that y2 = 4ax is a parabola!

    Journey’s End

    As we wrap up this series on the conic sections, we can observe that we have covered quite a bit of ground. We have looked at the geometric definition of a conic section as the curve formed when a cone is sectioned by a plane. We have also defined it in terms of coordinate geometry, using the focus, directrix, and eccentricity. We have derived the equations for the conic sections in standard form and have also derived the parametric coordinates of any point on the conic section. We have obtained the equations of the tangents at a point on the conic section in terms of Cartesian coordinates and parametric coordinates. We also introduced the idea of the chord of contact, the pole, and polar of a point in relation to a conic section.

    Along the way, we introduced the idea of degeneracy for a conic section and saw that the circle was a degenerate ellipse with eccentricity of zero. I defended the idea of including the pair of lines among the conic sections as a degenerate hyperbola with an infinite eccentricity. In this context, we looked at the conditions under which the general second degree equation in two variables represents a pair of lines. Then we obtained the point of intersection of a pair of lines and conditions for perpendicularity and parallelism of a pair of lines. Finally, we looked at the idea of homogenization, which we used to prove an important property of the parabola.

    Unfortunately, our journey into the world of the conic sections has come to an end. I hope it has been as enjoyable for you as it has been for me. I will be taking a break for a week and will resume on 5 December 2025. Till then, stay eccentric!

  • A Mathematical Straight Couple

    Making the Case

    We are in the middle of a series of posts on conic sections. After the introductory post, Finding Refreshment in the Desert, I devoted three posts to the circle and two each to the parabola, ellipse, and hyperbola. In the previous post I had mentioned a degenerate conic section known as a pair of lines. I know this doesn’t have the same kind of ring to it as do the others. But then, this is a degenerate conic section we are talking about. Many mathematicians do not consider it legitimate to include it among the regular conic sections. However, if we have included the circle, which is a degenerate ellipse, then I do not see why a degenerate hyperbola should not be included! Further, if a conic section is the curve obtained by sectioning a cone with a plane, then we can easily recognize that sectioning the cone through its apex will yield a shape that is actually two lines that intersect at the apex. Hence, I hold the view that the pair of lines is a valid conic section and should be studied under that umbrella.

    Joint Equation of Lines Through the Origin

    Our study of the pair of lines begins with a single line through the origin. Its equation will be y = mx. This means that m = y/x, which gives us the slope/gradient on the line. We can also write this as y – mx = 0. Any point on the line will satisfy its equation. We make the following observations: 1. the line passes through the origin; 2. the slope/gradient of the line is m.

    Now consider two lines y – m1x = 0 and y – m2x = 0. Let us consider the equation (y – m1x)(y – m2x) = 0. The represents the product of two numbers given by y – m1x and y – m2x. The equation states that this product is zero. However, the product of numbers can be zero if and only if one of the numbers is zero. This must mean that either y – m1x = 0, meaning that we have a point on the first line, or y – m2x = 0, meaning that we have a point on the second line. In other words the equation (y – m1x)(y – m2x) = 0 represents points that lie on either of the two constitutive lines. We call this equation the joint equation of the lines.

    If we expand the product, we will obtain a second degree equation that has the form

    If we divide this equation by x2, we will obtain

    Now recall that m = y/x. Hence, replacing y/x with m we get

    This is a quadratic equation in m and should have two solutions m1 and m2. From what we know about the sum and product of the roots of quadratic equations, these solutions satisfy the two equations

    So we have y/x = m1 or y/x = m2, both of which are straight lines through the origin. It follows that every second degree homogeneous equation in two variables represents the joint equation of two straight lines through the origin.

    Angle between the Lines ax2 + 2hxy + by2 = 0

    Let the lines make angles of θ1 and θ2 with the positive x-axis. Then m1 = tan⁡ θ1 and m2 = tan⁡ θ2, where

    If θ is the acute angle between the lines then

    Now, we know that

    Using the equations for the sum and product of the slopes/gradients we get

    Nature of the lines ax2 + 2hxy + by2 = 0

    We just derived

    From this we can conclude that, if h2 – ab > 0, we have two real and distinct lines, if h2 – ab = 0, we have two real and coincident lines, and if h2 – ab < 0, we have two imaginary and distinct lines. Further, if a + b = 0, we have two real and distinct perpendicular lines. It is crucial to note that in all cases the two lines pass through the origin – a real point! This is because zero is both real and imaginary!

    Joint Equation of General Pair of Lines

    Of course, we have only considered lines through the origin. What about lines in general? Let us consider the two lines

    Let P(α, β) be a point on either of the lines. Then either

    This means that the product

    Hence, the equation

    represents points that lie on either of the two lines. In other words, it is the joint equation of the two lines. Expanding the brackets we get

    This is of the form

    which is a general second degree equation in two variables. However, while this means that the joint equation of two lines is always a second degree equation in two variables, it does not mean that every general second degree equation in two variables represents a pair of lines.

    Now recall, from Finding Refreshment in the Desert, that the general second degree equation in two variables was obtained by sectioning a cone with a plane. If the pair of lines is a genuine conic section, then its equation should also satisfy the condition of being a general second degree equation in two variables. Of course, the equation

    satisfies the condition with g = f = c = 0. However, this represents only straight lines through the origin. What about straight lines that do not pass through the origin?

    If the general second degree equation in two variables represents a pair of lines then the LHS of the equation can be factorized into two linear factors. If we treat the equation as a quadratic equation in x we obtain the solutions for x as

    If we have two linear factors then the quantity under the root must be a perfect square. The quantity under the root is

    For this to be a perfect square, its discriminant should be zero. Hence,

    When the above condition is satisfied, the general second degree equation in two variables represents a pair of lines. Hence, we can conclude that the pair of lines is a genuine conic section since the general equation representing a conic section under certain conditions represents a pair of lines. What about when the condition is not satisfied? What can we say then?

    Conditions on ∆ and h2 – ab

    Let us define

    We have seen that, when ∆ = 0 the general second degree equation in two variables represents a pair of lines. We also saw earlier, the conditions for real and imaginary pairs of lines. We can collate this along with what we know about the other conic sections to obtain the table below.

    Of course, it will not do for me to just give you the final conclusions. We have derived the conditions for ∆ in this post. However, the conditions for the other conic sections can also be derived. Since this does not directly relate to the pair of lines, and because the derivation involves some serious algebraic juggling, those who are curious can view the derivation here.

    The Way Ahead

    I think in this post I have made the case for considering the pair of lines as a genuine conic section. I do not know why they are often excluded from the study. However, if the circle, which is a degenerate ellipse with eccentricity of zero, is included, why should the pair of lines, which is a degenerate hyperbola with and infinitely large eccentricity, be excluded? I think it is only because we are familiar with the circle from early years in school and are never taught to consider multiple lines together without aiming to solve them simultaneously to obtain their points of intersection. And since that is something that is often taught, in the next post we will consider how to use the general second degree equation in two variables to obtain the point of intersection (if it exists) of the two lines. We will also look at a procedure known as homogenization. Curious? Stay tuned!

  • Contacting Hyperbolic Polarity

    Brief Recapitulation

    In this post we continue with the series on conic sections. In the previous post, A Mathematical Exaggeration, we introduced the hyperbola and derived the equation of the hyperbola in standard form. We also obtained the equations for the conjugate hyperbola, the auxiliary circle of the hyperbola and the equation of the tangent at a point on the hyperbola and the condition for tangency in terms of the gradient (m) of the tangent. In this post, we will explore the chord of contact and the polar with respect to the hyperbola.

    Chord of Contact

    Consider the hyperbola

    Recall that the equation of the tangent at the point (x1, y1) is

    Let P(x1, y1) be a point outside the hyperbola

    From point P tangents are drawn to touch the hyperbola at Q(x2, y2) and R(x3, y3). This is shown in the figure below.

    Since PQ is the tangent at Q, its equation must be

    Since P lies on this tangent, its coordinates must satisfy the above equation. Hence,

    Since PR is the tangent at R, its equation must be

    Since P lies on this tangent, its coordinates must satisfy the above equation. Hence,

    It follows then that Q(x2, y2) and R(x3, y3) satisfy the equation

    Since two points determine a straight line, this must be the equation of the chord of contact QR.

    Pole and Polar

    Having derived the equation of the chord of contact, let us proceed to derive the equation of the polar of a point with respect to the hyperbola. Let P(x1, y1) be a fixed point placed anywhere relative to the hyperbola. A line through P is free to pivot about P. In general, this line will cut the hyperbola at two points. Suppose these points are Q(x2, y2) and R(x3, y3). Tangents are drawn to the hyperbola at the points Q and R. Let these tangents intersect at S(h, k). This is shown in the figure below.

    As the line through P pivots about P, the points of intersection of the line with the hyperbola will change. Hence, Q and R will move. This means that the tangents at Q and R will move. Hence, the point of intersection, S, of the two tangents will move. Then the locus of S is known as the polar of P wrt the hyperbola and P is known as the pole of the polar with respect to the hyperbola. It can be seen from the diagram that QR is the chord of contact of S wrt the hyperbola. Hence, its equation should be

    However, P lies on QR. Hence, its coordinates should satisfy the above equation. This gives us

    Generalizing, we obtain the locus of S to be

    As just obtained, the polar of the point P(x1, y1) is

    The point P(x1, y1) may be anywhere in relation to the hyperbola for a polar to exist. If the point P lies outside the hyperbola, the portion of it that lies inside the hyperbola is the chord of contact while the portion that lies outside the hyperbola represents the set of points actually reached by the polar. If the point P lies inside the hyperbola, the entire line represents the set of points actually reached by the polar for then there would be no chord of contact.

    If you wish to see how the chord of contact, pole and polar relate to a hyperbola, you can click here for a Geogebra App that I created. There are two checkboxes on the top left. The upper one displays or hides the asymptotes of the hyperbola. More on this shortly. With the lower one unchecked, you can see how the chord of contact varies. Just click on the point P and drag it around! If you check the lower checkbox on the top left, you can explore how the pole and polar vary. Again, just click on the point P and drag it around to position the pole. I have made the line rotate automatically. This changes the positions of Q and R automatically and, therefore, the bright pink tangents at those two points. The point of intersection of these two tangents will move, tracing the green line, which is the polar.

    The Asymptotes of a Hyperbola

    An asymptote is a ‘tangent at infinity’ to a curve. In other words, as either the x coordinate or y coordinate or both increase in magnitude without bound, the curve gets closer and closer to the asymptote. While the parabola and hyperbola are both unbounded figures, only the hyperbola has asymptotes. For the hyperbolas

    The lines

    are the asymptotes. These are shown in the figure below

    The ‘Inside’ of the Hyperbola

    Now, observe the figure we have been using for the chord of contact and pole and polar.

    By definition, a real chord of contact can be drawn only from a point outside the curve. Also, by definition, a chord must lie inside the curve. The chord of contact and the polar of  wrt the hyperbola is the green line. Since we are able to draw real tangents  PQ and PR, it follows that the point P lies outside the hyperbola but also that  lies inside the hyperbola! This is confusing so let us approach this differently.

    The asymptotes divide the two dimensional space into 4 regions designated as shown below.


    When the point P lies in a region H1, the points of tangency will both be on the right branch of the hyperbola and the chord of contact will be ‘inside’ the right branch. When the point P lies in a region H3, the points of tangency will both be on the left branch of the hyperbola and the chord of contact will be ‘inside’ the left branch. When the point P lies in a region H2 or H4, one point of tangency will both be on the right branch and the other on the left branch and the chord of contact will be ‘inside’ the two branches. Hence, the ‘inside’ of the hyperbola is a contextual idea that depends on where in relation to the asymptotes we place the point from which we desire to obtain the chord of contact.

    The Hyperbola and Its Asymptotes

    If we looked at the hyperbola and its asymptotes, we can see that the two asymptotes look much like a very ‘austere’ hyperbola with the two branches actually touching each other and the curved contours of the hyperbola becoming transformed into straight lines. It should come as no surprise, then, that, just as the circle is a degenerate form of the ellipse, the pair of lines that form the two asymptotes is a degenerate form of the hyperbola. Specifically, the pair of lines is a hyperbola with an eccentricity of infinity, much like the circle is an ellipse with eccentricity of zero. Indeed, the pair of lines has remarkable properties in and of itself, much like the circle did. And we will turn to that, the final ‘conic section, in the next post.

  • A Mathematical Exaggeration

    The Riddle of the Name

    In this post we continue with our study of the conic sections. So far we have looked at the circle, the parabola, and the ellipse. Now we turn to the final non-degenerate conic section – the hyperbola. When we hear the word ‘hyperbola’, some of us may think of the English word ‘hyperbole’ and wonder if the two are related. And indeed they are. The word ‘hyperbole’ is “the use of extreme exaggeration for emphasis or effect.” If we consider the issue of the eccentricity of a conic section, we can see that the circle is one with an eccentricity of 0. The ellipse has an eccentricity between 0 and 1, while the eccentricity of a parabola is 1. Now, the circle and the ellipse are closed figures. However, as the eccentricity becomes larger and approaches 1, the ellipse becomes more elongated and flat. a parabola could be considered a degenerate ellipse where the second focus is actually at infinity, thereby leading to an open figure rather than a closed on. If we choose to ‘throw’ (Greek ballein) in ‘excess’ (Greek hyper) of this limitation, we get a hyperbola. Hence, if we exaggerate the eccentricity beyond that of the parabola, we get a hyperbola. In other words, a hyperbola is a conic section which has an eccentricity that is greater than 1.

    Just to refresh our minds about the conic sections, consider the figure above. If the cone is sectioned perpendicular to the axis of the cone, we get a circle. Now, if the sectioning plane is rotated such that the angle made is between being perpendicular to the axis and parallel to the surface of the cone, we get an ellipse. When the sectioning is parallel to the surface of the cone, we get a parabola. And if the sectioning angle is greater than that of the surface of the cone, then we get a hyperbola.

    Definition and Basics of a Hyperbola

    A hyperbola is the locus of a point which moves such that the ratio of its distance from a fixed point to its distance from a fixed line is a constant greater than one. The fixed point is known as the focus, S and the fixed line is known as the directrix, L. The fixed ratio is the eccentricity, e∈(1,∞), for a hyperbola. A hyperbola actually has two foci, S and S‘, and two directrices, L and L‘, that form two pairs S–L and S‘-L‘ which both satisfy the definition of the hyperbola. The transverse axis of the hyperbola is the line joining the two foci. The vertices of the hyperbola are the points on the hyperbola and its transverse axis. One vertex, A, lies between S and L and the other vertex, A‘, lies between S‘ and L‘. The length of AA‘ is the length of the transverse axis of the hyperbola. The center, O, of the hyperbola is the midpoint of SS‘ (also the midpoint of AA‘). The line perpendicular to the transverse axis through the center of the hyperbola is the conjugate axis. No points lie on the hyperbola and on its conjugate axis. Let us derive the equation of a hyperbola in standard form.

    The Equation of the Hyperbola

    Consider a point, S, given to be the focus of a hyperbola and a line, L = 0, given to be the directrix of the hyperbola. Let e be the eccentricity of the hyperbola. Drop a perpendicular from S onto L and let the foot of the perpendicular be X. Let A divide SX internally in the ratio e:1. Hence, A lies on the hyperbola. Let A‘ divide SX externally in the ratio e:1. Hence, A‘ lies on the hyperbola. Let AA‘ be the transverse axis of the hyperbola. Let O be the midpoint of AA‘, which we take to be the origin of our coordinate system. This is shown in the figure below.

    Let OA produced form the positive x axis and let the length AA‘ be 2a. Hence, the coordinates of A are (a, 0) and of A‘ are (-a, 0). Let the coordinates of X be (p, 0) and of S be (q, 0). Then by section formula for internal and external division in the ratio e:1.

    This gives ep + q=a(e + 1) = ae + a and ep – q=-a(e – 1)=-ae + a. Adding the two equations we get p = a/e and subtracting we get q = ae. Hence the coordinates of X are (a/e, 0) and of S are (ae, 0) and line L is x – a/e = 0.

    Let P(x, y) be any point on the hyperbola. Let M be the foot of the perpendicular from P onto L. Then the coordinates of M are (a/e, y). Then by definition of the hyperbola SP = e⋅PM.

    Let a2(e2 – 1) = b2. Then the equation of the hyperbola becomes

    We can observe that the figure is symmetric about both axes. Hence the other focus is S‘ (-ae, 0) and the other directrix is L‘ = x + a/e = 0.

    Conjugate Hyperbolas and the Auxiliary Circle

    There are two configurations of a hyperbola in standard form depending on whether the constant on the right is 1 or -1, known as conjugate hyperbolas.

    Shown in the figure above are the conjugate hyperbolas

    The circle with the transverse axis as its diameter is known as the auxiliary circle. For the hyperbola x2/a2 – y2/b2 = 1 its equation is x2 + y2 = a2. For the hyperbola x2/a2 – y2/b2 = -1 its equation is x2 + y2 = b2.

    Parametric Coordinates

    Consider a point P on the hyperbola x2/a2 – y2/b2 = 1. See the figure on the left below. Let the perpendicular to the transverse axis of the hyperbola (here the x axis) meet the transverse axis at the point Q. Let R be the point of contact of the tangent from Q onto the auxiliary circle. θ, the eccentric angle of P is the angle made by radius OR with the positive x axis. Now OR = a ⇒ OQ= a sec⁡θ ⇒y =b tan⁡θ ⇒ P ≡ (a sec⁡θ, b tan⁡θ), which are the parametric coordinates of the point on the hyperbola with eccentric angle of θ.

    For the hyperbola x2/a2 – y2/b2 = -1, the parametric coordinates are (a cotθ, b cosecθ), where θ is the angle made by the radius formed by the point on the auxiliary circle corresponding to the point P, as shown in the figure to the right above.

    Tangent at the Point (x1, y1)

    Let us derive the equation of the tangent at a point (x1, y1) on the hyperbola. Consider the hyperbola

    Differentiating this equation with respect to x we get

    If (x1, y1) is on the hyperbola then

    and at the point (x1, y1)

    Hence, the equation of the tangent would be

    Hence, the equation of the tangent at (x1, y1) is

    Tangent at the Point (a sec⁡θ, b tan⁡θ)

    Since the point (a secθ, b tanθ) satisfies the equation of the hyperbola, the equation of the tangent at the parametric point will be

    Condition for Tangency

    Consider the line y = mx + c and the hyperbola

    Solving the two equations we get

    If the line is a tangent to the hyperbola, the above quadratic equation should have a repeated root, meaning that its discriminant must be zero. Hence,

    Hence, we can conclude that the line

    is always a tangent to the hyperbola

    The Return of the Riddle

    We have seen some interesting things about the hyperbola. There are many similarities between it and especially the ellipse. For example, the equations in standard form only differ in the sign (+ or -) that separates the terms. Similarly, both figures have a well defined auxiliary circle. Further, both conics have trigonometric parametric coordinates, though obtaining them for the hyperbola took a little more creativity than for the ellipse. This was primarily because the ellipse is easily visualized as a squashed circle, while the hyperbola, with its twin branches makes visualization a little more difficult. Indeed, it is because the eccentricity is ‘exaggerated’ (i.e. greater than 1), that the distance from the focus expands more rapidly than the distance from the directrix, making it impossible for the point to come back down toward the transverse axis to close the figure. In other words, the value of the eccentricity leads to the ‘excess’ (hyper) throwing (ballein) of the point, thereby yielding the hyperbola.

    I will be taking a break next week. Hence, the next post will be on 7 November. In that post we will look at some more properties of the hyperbola, including the chord of contact and the polar. If you have forgotten what those words mean, please refer to Contacting Circular Polarity. And may you not experience any exaggerated hurling! Make of that what you will!

  • Contacting Elliptical Polarity

    Brief Recapitulation

    In this post we continue with the series on conic sections. In the previous post, A Mathematical Shortfall, we introduced the ellipse and derived the equation of the ellipse in standard form. We also obtained the equation for the auxiliary circle of the ellipse and the equation of the tangent at a point on the ellipse and the condition for tangency in terms of the gradient (m) of the tangent. In this post, we will explore the chord of contact and the polar with respect to the ellipse.

    Chord of Contact

    Consider the ellipse

    Recall that the equation of the tangent at the point (x1, y1) is

    Let P(x1, y1) be a point outside the ellipse

    From point P tangents are drawn to touch the ellipse at Q(x2, y2) and R(x3, y3). This is shown in the figure below.


    Since PQ is the tangent at Q, its equation must be

    Since P lies on this tangent, its coordinates must satisfy the above equation. Hence,

    Since PR is the tangent at R, its equation must be

    Since P lies on this tangent, its coordinates must satisfy the above equation. Hence,

    It follows then that Q(x2, y2) and R(x3, y3) satisfy the equation

    Since two points determine a straight line, this must be the equation of the chord of contact QR.

    Pole and Polar

    Having derived the equation of the chord of contact, let us proceed to derive the equation of the polar of a point with respect to the ellipse. Let P(x1, y1) be a fixed point placed anywhere relative to the ellipse. A line through P is free to pivot about P. In general, this line will cut the ellipse at two points. Suppose these points are Q(x2, y2) and R(x3, y3). Tangents are drawn to the ellipse at the points Q and R. Let these tangents intersect at S(h, k). This is shown in the figure below.

    As the line through P pivots about P, the points of intersection of the line with the ellipse will change. Hence, Q and R will move. This means that the tangents at Q and R will move. Hence, the point of intersection, S, of the two tangents will move. Then the locus of S is known as the polar of P wrt the ellipse and P is known as the pole of the polar with respect to the ellipse. It can be seen from the diagram that QR is the chord of contact of S wrt the ellipse. Hence, its equation should be

    However, P lies on QR. Hence, its coordinates should satisfy the above equation. This gives us

    Generalizing, we obtain the locus of S to be

    As just obtained, the polar of the point P(x1, y1) is

    The point P(x1, y1) may be anywhere in relation to the ellipse for a polar to exist. If the point P lies outside the ellipse, the portion of it that lies inside the ellipse is the chord of contact while the portion that lies outside the ellipse represents the set of points actually reached by the polar. If the point P lies inside the ellipse, the entire line represents the set of points actually reached by the polar for then there would be no chord of contact.

    If you wish to see how the chord of contact, pole and polar relate to an ellipse, you can click here for a Geogebra App that I created. With the checkbox on the top left unchecked, you can see how the chord of contact varies. Just click on the point P and drag it around! If you check the checkbox on the top left, you can explore how the pole and polar vary. Again, just click on the point P and drag it around to position the pole. I have made the line rotate automatically. This changes the positions of Q and R automatically and, therefore, the bright pink tangents at those two points. The point of intersection of these two tangents will move, tracing the green line, which is the polar.

    From the applet you will notice that the chord of contact technically only represents the part of the line that is inside the ellipse which the polar represents the part that is outside the ellipse. And when P is on the parabola, the chord of contact degenerates to the point P, meaning that the tangent itself is the polar. This is why all three features of the ellipse, as in the case of all the conic sections, share the same equation.

    The Director Circle

    The ellipse also has another interesting feature known as the director circle. It is the collection of all points from which tangents drawn to the ellipse are at right angles to each other. Recall the condition for tangency. Given the ellipse

    the line

    is always a tangent to the ellipse. Consider a point P(h, k). Let the tangent

    pass through P. Then the coordinates of P will satisfy the equation of the tangent.

    This is a quadratic equation in m and if the tangents through P intersect at right angles then the product of the slopes of the tangents would be -1. From what we know concerning quadratic equations, this means that the constant term divided by the coefficient of x2 must be -1. Hence,

    Generalizing we get the equation of the director circle is

    Hence, from any point on the above circle the two tangents drawn to the ellipse

    will be at right angles to each other.

    The Ellipse and the Circle

    If we compare the the ellipse and the circle we will see that setting b = a in all the equations related to the ellipse transforms the equations to those related to the circle. This is why the ellipse and the circle are related and the latter is said to be the degenerate form of the former. In particular, with respect to the previous figure, in a circle, the two foci, F1 and F2, are merged into the center of the circle. Hence, in a very real sense, the distance between the foci is a measure of the eccentricity of the ellipse. Recall that, when we derived the equation for the ellipse, we obtained the coordinates of the foci to be (ae, 0) and (-ae, 0). This means that the distance between the foci is 2ae, validating the intuition that the distance between the foci is linked to the eccentricity of the ellipse.

    In these two posts we have briefly studied the ellipse. In the next two posts we will deal with the hyperbola, the last of the non-degenerate conic sections. Till then, I hope you have a week with no shortfalls.

  • A Mathematical Shortfall

    The Riddle of the Name

    When I was in school, one of my English teachers (unfortunately I do not recall which one) taught us about the punctuation mark known as the ellipsis. An ellipsis is three dots (…) that indicate something is missing. It could a lack of words, such as…, well…, you see…, indicating a person doesn’t have the words to fill up his/her thoughts clearly. Or it could indicate words that are actually missing, such a words in an ancient document for which only a fragment exists. An ellipse, however, is a conic section, a geometric figure. But consider the figure below, which indicates why the conic sections are called ‘conic sections’.

    The circle represents a sectioning of the cone at right angles to the axis of the cone (the vertical line in the figure above). The parabola represented a sectioning that is parallel to the edge of the cone. An ellipse represents a ‘falling short’, in that its sectioning is neither at right angles to the axis nor parallel to the edge. And guess what? Both words ‘ellipsis’ and ‘ellipse’ come from the Greek word ἔλλειψις (élleipsis), meaning ‘fall short’ or ‘leave out’.

    Defining the Ellipse

    As we continue with this series on conic sections, we reach the ellipse. Recall that a conic section is defined as the set of points such that the ratio of the distance of the point to a fixed point, called the focus, to its distance from a fixed line, called the directrix, is a constant. The constant ratio is called the eccentricity of the conic section. We saw that the parabola has an eccentricity equal to 1. An ellipse has an eccentricity that is between 0 and 1.

    The Equation of the Ellipse

    Let us begin our exploration of the ellipse by deriving its equation in standard form as we did for the parabola. Consider a point, S, given to be the focus of an ellipse and a line, L = 0, given to be the directrix of the ellipse. Let e be the eccentricity of the ellipse. Drop a perpendicular from S onto L and let the foot of the perpendicular be X. Let A divide SX internally in the ratio e:1. Hence, by definition of the ellipse A lies on the ellipse. Let A‘ divide SX externally in the ratio e:1. Hence, by definition of the ellipse A‘ lies on the ellipse. This is shown in the figure below.

    Now, the ellipse has two axes of symmetry, one longer than the other. The longer one is called the major axis and the shorter one the minor axis. Let AA‘ be the major axis of the ellipse. Let O be the midpoint of the ellipse, which we take to be the origin. Let OA produced form the positive x axis. Let the length AA‘ be 2a. Hence, the coordinates of A are (a, 0) and of A‘ are (-a, 0). Let the coordinates of X be (p, 0) and the coordinates of S be (q, 0). Then by section formula for internal and external division in the ratio e:1.

    This gives ep + q = a(e + 1) = ae + a and ep – q = –a(e – 1) = –ae + a.
    Adding the two equations we get p = a/e and subtracting them we get q = ae. Hence the coordinates of X are (a/e, 0) and the coordinates of S are (ae, 0). Also, the equation of L is x – a/e = 0.
    Let P(x, y) be any point on the ellipse. Let M be the foot of the perpendicular from P onto L. Then the coordinates of M are (a/e, y). Then by definition of the ellipse SP = e⋅PM.

    Since this equation is a little cumbersome, let us make the substitution

    Hence, the equation of the ellipse becomes

    We can observe that the figure is symmetric about both axes. Hence, there will be a second focus and a second directrix. The other focus is S‘ (-ae, 0) and the other directrix is L‘ = x + a/e = 0.
    Now, the chord through the focus and perpendicular to the major axis is called the latus rectum. Since, the ellipse has two foci, it will also have two latera recta, whose equations are x – ae = 0 and x + ae = 0.
    Also, putting x = 0 we get the coordinates of B and B‘, where the ellipse cuts the y axis, to be (0, b) and (0,-b) respectively. Now, here we have assumed that a > b. However, even with a < b we will get the same equation. However, in this case

    The Auxiliary Circle

    Now, the circle described with the major axis of the ellipse as its diameter is known as the auxiliary circle. For the ellipse with a > b (figure on the left above) its equation is x2 + y2= a2. For the ellipse with a < b (figure on the right above) its equation is x2 + y2= b2.

    Parametric Coordinates

    Consider the ellipse x2/a2 + y2/b2 = 1, a > b. Consider a point P on the ellipse. Let the perpendicular to the major axis of the ellipse (here the x axis) meet the auxiliary circle at the point Q(a cos⁡θ, a sin⁡θ ). It is clear then that the x coordinate of P is a cos⁡θ. Substituting this for the x coordinate in the equation of the ellipse we get that the y coordinate is b sin⁡θ. Hence the coordinates of P are (a cos⁡θ, b sin⁡θ), where θ is called the eccentric angle of the point P.
    Hence for the ellipse x2/a2 + y2/b2 = 1, a > b, the parametric coordinates are (a cos⁡θ, b sin⁡θ), where θ is the angle made by the radius formed by the point on the auxiliary circle corresponding to the point P (left figure below). Similarly, for the ellipse x2/a2 + y2/b2 = 1, a < b, the parametric coordinates are (a cos⁡θ, b sin⁡θ), where θ is the angle made by the radius formed by the point on the auxiliary circle corresponding to the point P (right figure below).

    Equation of the Tangent

    Let us proceed to obtain the equation of the tangent at a point on the ellipse. Consider the ellipse x2/a2 + y2/b2 = 1. Differentiating this equation with respect to x we get

    Hence, at the point (x1, y1)

    Hence, the equation of the tangent would be

    We have already shown that the point (a cos⁡θ, b sin⁡θ ) satisfies the equation x2/a2 + y2/b2 = 1. Hence, the point always lies on the ellipse. Substituting these coordinates in the tangent equation just derived we get

    Condition for Tangency

    Consider the line y = mx + c and the ellipse x2/a2 + y2/b2 = 1. Solving the two equations we get

    If the line is a tangent to the ellipse, the above quadratic equation should have a repeated root which means its discriminant must be zero. Hence,

    So the line

    will always be a tangent to the ellipsex2/a2 + y2/b2 = 1

    Return of the Riddle

    What we have seen is that the ellipse has similarities to both the parabola and the circle. Its parametric coordinates are very similar to those of the circle. Hence, the condition for tangency is also quite similar to what we saw for the circle. However, given that the circle is a degenerate conic section, we could not derive the equation of the ellipse as we did the equation of the circle. Rather, we had to use the focus and directrix, as we did with the parabola, to obtain the equation of the ellipse. Probably, the fact that the ‘ellipse’ is a ‘shortfall’ between the parabola and the circle is why some of the results related to it use methods similar to those used for the parabola while other results use methods similar to those used in the context of the circle.

    In the next post we will look at some more properties of the ellipse, including the chord of contact and the polar. If you have forgotten what those words mean, please refer to Contacting Circular Polarity. And may you not experience any shortfall, except perhaps to be bifocal and hence bidirectional. Make of that what you will!

  • Contacting Parabolic Polarity

    Contacting Parabolic Polarity

    Brief Recapitulation

    In the previous post, titled Mathematical Parables, I introduced us to the parabola. Just to jog your memory, a conic section is a a geometric figure such that, for any point on the conic section, the ratio of its distance from a fixed point, called the focus, to its distance from a fixed line, called the directrix, is a constant, known as its eccentricity. A parabola is a conic section which has an eccentricity equal to 1.

    In the previous post, we derived the equation y2 = 4ax as that of the parabola in standard form, obtained the parametric coordinates (at2,2at) for a general point on the parabola, derived the equation yy1 = 2a(x + x1) of the tangent at a point (x1, y1) on the parabola, and the condition c = a/m under which the line y = mx + c is a tangent to the parabola. In this post, we will deal with the chord of contact and the polar of a point with respect to the parabola.

    Chord of Contact

    Let P(x1, y1) be a point outside the parabola y2 = 4ax. From point P tangents are drawn to touch the parabola at Q(x2, y2) and R(x3, y3). This is shown in the figure below.

    Since PQ is the tangent at Q, its equation must be yy2 = 2a(x + x2). Since P lies on this tangent, its coordinates must satisfy this equation. Hence, y1y2 = 2a(x1 +x2)

    Since PR is the tangent at R, its equation must be yy3 = 2a(x + x3). Since P lies on this tangent, its coordinates must satisfy this equation. Hence, y1y3 = 2a(x1 + x3)

    It follows then that Q(x2, y2) and R(x3, y3 ) satisfy the equation yy1 = 2a(x + x1) and since two points determine a straight line, the equation of the chord of contact QR must be yy1 = 2a(x + x1).

    Pole and Polar

    Having derived the equation of the chord of contact of the point (x1, y1) with respect to the parabola y2 = 4ax let us derive an equation for the polar of the same point with respect to the parabola. Let P(x1, y1) be a fixed point placed anywhere relative to the parabola y2 = 4ax. A line through P is free to pivot about P. In general, this line will cut the parabola at two points. Suppose these points are Q(x2, y2) and R(x3, y3). Tangents are drawn to the parabola at the points Q and R. Let these tangents intersect at S(h, k).This is shown in the figure below.

    It is clear that at the line pivots about P, the points of intersection, Q and R, of the line with the parabola will vary. This means that the tangents at Q and R will vary. Hence, the point S where the two tangents intersect will move. The locus (fanciful name for ‘path’) of S is known as the polar of P with respect to the parabola and P is known as the pole of the polar with respect to the parabola. It can be seen from the diagram that QR is the chord of contact of S wrt the parabola. Hence, its equation should be yk = 2a(x + h). However, P lies on QR. Hence, y1k = 2a(x1 + h). Generalizing, we obtain the locus of S to be yy1 = 2a(x + x1).

    Tangent, Chord of Contact, and Polar

    Once again, as in the case with the circle, we can reach the observation that the same equation represents the tangent at the point (x1, y1), the chord of contact of the point (x1, y1), and the polar of the point (x1, y1) with respect to the parabola y2 = 4ax. In this case, the equation is yy1 = 2a(x + x1). Once again, it is the context within which the equation is used that tells us what the equation represents.

    If you wish to see how the chord of contact, pole and polar relate to a parabola, you can click here for a Geogebra App that I created. With the checkbox on the top left unchecked, you can see how the chord of contact varies. Just click on the point P and drag it around! If you check the checkbox on the top left, you can explore how the pole and polar vary. Again, just click on the point P and drag it around to position the pole. Then move the slider for α to change the orientation of the line through P. This will change the positions of Q and R and, therefore, the bright pink tangents at those two points. The point of intersection of these two tangents will move, tracing the green line, which is the polar.

    Once again, from the applet you will notice that the chord of contact technically only represents the part of the line that is inside the parabola which the polar represents the part that is outside the parabola. And when P is on the parabola, the chord of contact degenerates to the point P, meaning that the tangent itself is the polar. This is why all three features of the parabola, as in the case of all the conic sections, share the same equation.

    A Bone Picked

    In the post Contacting Circular Polarity I had expressed my dismay that concepts such as chord of contact and pole and polar are not taught in most (any?) contemporary curriculums. However, just the past week, I came across a post on LinkedIn in which the author was lamenting the return to archaic topics like analytical geometry in India’s mathematical education. I wanted to refer to that post here. However, the Amazon delivery guy came when I was reading the post on my phone and when I returned to LinkedIn I was unable to locate the post. The author claimed that a number of mathematics teachers or professors had voiced their disapproval of this return to such archaic concepts. The author argued that we should instead be focusing on AI and Data Science. Given that I have embarked on this series on analytical geometry, I had to address the issue raised by the post. If any reader can locate the post, please share it with me.

    As mentioned in earlier posts, analytical geometry helps us link geometry, which can be quite abstract, to algebra, which is less so. This allows us to obtain a geometric interpretation of algebraic equations and an algebraic interpretation of geometric phenomena, both of which help link disparate sides of mathematical knowledge. While algebra is very left brain focused, geometry is more right brain focused. This gives the student an ability to visualize arcane algebraic equations from a geometric perspective and to rationalize geometric insights from an algebraic perspective. Hence, those who seem to think that analytical geometry is archaic have failed to appreciate what this branch of mathematics brings to the table.

    However, in very few other fields do we find a rejection of the archaic. Physics students learn about Newton’s corpuscular theory of light and the ether theory of light even though both have been conclusively disproven. Chemistry students still study Dalton’s and Thomson’s atomic models even though they are now demonstrably wrong. I have mentioned four theories, each proven to be incorrect, that science students still learn to this day. And people complain about analytical geometry not because it is wrong (it is not) but because it is archaic? I wonder how many English teachers would say that English students should not study Shakespeare or Blake, Dickenson or Brontë because they are archaic and we have contemporary poets to study!

    Since I am unable to locate the post, it may be that I have misrepresented the views of the author. In that case, I will be willing to apologize. Not for the stance I take concerning analytical geometry, but for misrepresenting him/her. My stance on analytical geometry remains unwavering. It is an essential part of mathematical education and probably the only relatively easy part of mathematics that is able to bridge the gap between the two hemispheres of the brain. I hope that, through this series on conic sections, I am able to communicate just this.

  • Mathematical Parables

    Mathematical Parables

    Starting the Parabolic Path

    Those who know me well, know that I have two main areas of interest. Since you are reading this blog, you probably know or have surmised that mathematics is one of those areas. If you do not yet know, the other area is the exploration of my Christian faith and to fulfill that interest I blog at Purposes Crossed. If the title of this post seems strange, it is because I have intentionally combined the two areas. Obviously, this is a mathematics focused blog. Hence, the first word. For the second word, we need to understand that a parable is a pedagogical tool used widely by many teachers in the past, including Jesus.

    But why do I add this word about parables? Well, a parable is a story in which there is one main point, one focal point, one could say. And as we turn to the second of the conic sections, the parabola, we realize that this geometric figure too has one focal point. Of course, we need to ask ourselves what a focal point would mean for a geometric figure. We will get to that later. However, before we begin our study of the parabola, it pays to realize that the Greek word for parable is parabolé (, which happens to be the word used by Euclid for the parabola. Hence, it is either a strange coincidence or an ongoing Greek joke that has lasted for two and a half millennia! Could it be that Euclid wanted us to see that the parabola is a conic section with a story? What could its story be? Or could it be that he wanted us to think of parables as strangely constructed word parabolas, whatever that could mean? Let us try to determine that.

    Profligate Oddballs

    We began the series on conic sections with a study of circles. There was much more we could have covered. However, I dealt with a few ideas that are not normally taught in high school curriculums around the world. As we turn from the circle to the other conic sections, namely the parabola, the ellipse, and the hyperbola, we need to introduce a new idea, the eccentricity of the conic section. We will see that the circle is simply a degenerate form of the ellipse. The previous two sentences should give you an idea of how mathematicians couch their jargon with tongue in cheek terms. I mean, we are talking about eccentric and degenerate geometric figures as though these inanimate squiggles on the page had personalities and moralities, as though they were oddballs and profligates! We will address the first of these terms, the eccentricity, shortly. The idea of the degeneracy of a conic section will have to wait till we discuss the ellipse.

    The Eccentricity

    All non-degenerate conic sections are defined by a point and a line. The point is said to be the focus and the line is called the directrix. For any general point on the conic section, the ratio of its distance from the focus to its distance from the directrix is a constant, which is called its eccentricity. The eccentricity of the parabola is 1. A conic section with eccentricity between 0 and 1 is an ellipse, with the circle having and eccentricity of 0. Finally, a conic section with eccentricity greater than 1 is a hyperbola. Let us proceed to derive the equation of a parabola in standard form.

    The Equation of the Parabola

    Consider a point, S, given to be the focus of a parabola and a line, L = 0, given to be the directrix of the parabola. Drop a perpendicular from S onto L and let the foot of the perpendicular be X. Let us define the distance SX to be 2a. Let A be the midpoint of SX. Then AS = AX = a, which by the definition of the parabola implies that A is on the parabola. Since A is the midpoint of the perpendicular from the focus to the directrix, A is said to be the vertex of the parabola. Let A be the origin of the coordinate system and let AS produced be the positive x axis. This is depicted below.

    Then the coordinates of A are (0,0) and of S are (a,0). The coordinates of X will be (-a,0). Hence, the equation of L would be L = x + a = 0. Let P(x, y) be any point on the parabola. Let M be the foot of the perpendicular from P onto L. Then the coordinates of M are (-a, y). Then by definition of the parabola SP = PM. This gives us

    Hence the equation of the parabola with axis along the x-axis, vertex at the origin, directrix formed by x + a = 0 and focus at (a, 0) is y2 = 4ax.

    Now the latus rectum is the chord through the focus and perpendicular to the axis. Hence, the equation of the latus rectum is x = a. Substituting x = a in the equation of the parabola gives y = ±2a. Hence, the endpoints of the latus rectum are (a, 2a) and (a, -2a), giving that the length of the latus rectum is 4a.

    The Riddle of the Parameter

    When we dealt with the circle, we said that the coordinates of any point on the circle of radius r and centered at the origin would be (r cos θ, r sin θ), where θ is the angle formed by the radius joining the point to the origin and the positive x-axis. In this case, θ is said to be a ‘parameter’ and the aforementioned coordinates are said to be the parametric coordinates.

    The parabola y2 = 4ax also can be described in terms of a parameter as follows. Suppose x = at2 and y = 2at. Since (2at)2 = 4a(at2 ), the point (at2,2at) lies on the parabola y2 = 4ax for all values of t, where t is a parameter. We will discover what the significance of t is shortly.

    Equation of the Tangent

    Before we get to that, let us obtain the equation of the tangent to the parabola at an arbitrary point (x1, y1) . Differentiating the equation y2 = 4ax we get

    Hence, the slope of the tangent at (x1, y1) is 2a/y1. This means that the equation of the tangent will be

    But (x1, y1) lies on the parabola. Hence,

    So, the equation of the tangent becomes

    Substituting the parametric coordinates x1 = at2 and y1 = 2at, we get

    Condition for Tangency

    We can also obtain the equation of the tangent in terms of the slope of the line. Let the line y = mx + c be tangent to the parabola y2 = 4ax. Solving the two equations we get

    Since the line is a tangent, the above quadratic equation should have a repeated root. Hence,

    The last equation is the condition under which the line y = mx + c will be tangent to the parabola y2 = 4ax. In other words, replacing c with a/m we obtain that the line

    is always a tangent to the parabola.

    Answering the Riddle

    However, if we compare this with the parametric equation of the tangent

    we can see that the two lines will be identical if t = 1/m. This tells us that the parameter t at a point on the parabola is nothing but the reciprocal of the slope of the tangent at that point. In other words, what seemed like a parameter introduced just to satisfy the equation of the parabola, actually turns out to have a deeper significance for the parabola itself. In other words, the parameter itself describes an essential property of the parabola. Or to put it in words with which I began this post, the parameter tells a story about the parabola. It is a parabolic way of describing the parabola.

    In the next post we will look at some more properties of the parabola, including the chord of contact and the polar. If you have forgotten what those words mean, please refer to Contacting Circular Polarity. Till then, stay focused! And remain profligate oddballs!

  • Circular Family Values

    Circular Family Values

    We are in the middle of a series on conic sections. We started with Finding Refreshment in the Desert, in which I introduced us to what a conic section is and demonstrated that the general second degree equation in two variables represents a conic section. After that, in Circular Tangentiality we obtained the general equation of a circle and derived the equation of the tangent to a circle at a point. Last week, in Contacting Circular Polarity, I introduced us to the concepts of the chord of contact and polar of a point with respect to a circle. And toward the end of the post I teased the reader with a mention of a ‘family of circles’. Whatever could that be?

    Recall that the equation of a circle with center at (p, q) and radius r is given by

    When this is expanded and rearranged, we get

    Recall also that the general second degree equation in two variables is

    We can see that the equation of the circle is similar to the general second degree equation in two variables with

    Hence, by convention we say that the general equation of a circle is

    with center at (-g, –f) and radius given by

    Now, suppose we are given the circles

    The radical axis of these two circles is obtained by equating C1 and C2. Hence,

    This is clearly the equation of a straight line and we can easily show that it is perpendicular to the line joining the centers of C1 and C2. Now consider the equation

    When λ = -1, this represents the radical axis. However, when λ ≠ -1, we can write the equation as

    Dividing the equation by 1 + λ we get

    This clearly is the equation of a circle. Hence, by varying the value of λ we can obtain different circles. Suppose we attempted to find the radical axis of this circle and C1. We would get

    If we focused on the coefficient of x we would obtain

    Hence, the equation of the radical axis would be

    If we divide this equation by λ/(1 + λ) we will get

    which is the radical axis of C1 and C2. In other words, any two members of this family of circles shares the same radical axis as the two circles used to generate the family.

    While the study of family of circles began, as is the case for many concepts related to conic sections, simply as an exercise of playing with the mathematics, this obscure concept has found applications in the study of paths of light and designs of tires and gears. This has been the case with many mathematical concepts that were developed without any practical applications in mind. However, as the study proceeds, quite often unforeseen applications become possible.

    There is much more that can be discussed related to the circle, even related to families of circles. However, that will take us really deep down this rabbit hole. Hence, we will leave the circle for now. In the next post, we will turn our focus to the parabola and see what properties this conic section has.