Mathematics Magazine Problem 1587 Solution.

    Problem. (Proposed by Kevin Ferland, Bloomsburg Univ., Bloomsberg, PA, and Florian Luca, Czech Academy of Sci., Prague, Czech Republic.) Can we use straightedge and compass to construct the foci (or focus) and directrices and asymptotes (if they exist) of a given ellipse, hyperbola or parabola?

    Solution. Yes! Let K be the nondegenerate conic, which has an interior made up of points that have no tangent lines or asymptotes towards K, and exterior that have. Let O be the center of K, if K is not a parabola. To save space we shall say "draw" instead of "construct", and "cut" instead of "intersect".

    Case 1: K is an ellipse or hyperbola. The first step is to draw the center of K. Draw any two parallel chords L₁ and L₂. Then the line joining their mid-points is called a conjugate diameter L₁' which will go through O as well as all midpoints of chords parallel to L₁ and L₂. Also if it cuts K, it will do so at points where the tangents are also parallel to L₁ and L₂. Choose another two parallel chords L₃ and L₄ not parallel to L₁. Then their conjugate diameter L₃' will cut L₁' at O.
     The axes can now be drawn. Draw a circle, center O, cutting K at 4 points. They will form a rectangle, so we can draw lines through O parallel to its sides, and those will be the axes.
    For the ellipse, the rest is easy: Since the axes intersect K we can get the major and minor radii of K, as a and b, say. Let X'OX be the major axis, cutting K at X' and X. With compass and ruler we now can draw the foci F' and F because OF = OF' = Ö(a²-b²). The directrices will be parallel to the minor axis, cutting the major axis at P', P where OP = a²/Ö(a²-b²) can be drawn.

    For the hyperbola, see Fig. 2. Let X'OX be the axis that cuts K at X' and X. Pick any point A on K, not on line XX'. Draw 2 chords L₁ and L₂ parallel to OA, with M and M' as their midpoints. Then the line MM' will go through O, will not cut K, which we shall call the conjugate diameter to OA. Draw the line XQR perpendicular to OX, cutting the lines OA, MM' at Q,R respectively. Draw the point S between Q and R so that XS = Ö(XQ · XR). Then it can be proved that the line OS is an asymptote. Reflecting it across OX will give the other asymptote. If OX = a and XS = b, then the focus F on OX can be drawn with compass and straightedge, because OF = Ö(a²+b²) . One directrix is the line parallel to XS cutting OX at P such that OP = a²/Ö(a²+b²). P can then be drawn with compass and straightedge. The other directrix is then its reflection across the minor axis.

    Case 2. K is a parabola. See Figure 3. Draw any 2 parallel chords, L₁ and L₂ with midpoints M, M'. Then the line MM' is parallel to the axis, which still has yet to be drawn. Line MM' will be called the diameter conjugate to L₁ or L₂.
    Next, we shall draw the focus F as follows: Draw 2 chords L₃ and L₄ perpendicular to each other, then draw the diameter L₃' conjugate to L₃ cutting K at T₃. Similarly draw the diameter L₄' conjugate to L₄ cutting K at T₄. Then the tangents at T₃ and T₄, being parallel to L₃ and L₄, will be perpendicular to each other, hence the chord T₃T₄ has to go through the focus F, i.e., is a focal chord. (This is a theorem in coordinate geometry.) By using another pair of perpendicular chords, we can draw another focal chord, cutting the first focal chord at F.
    Once we get F we can draw the axis of K as the line through F parallel to MM'. Let this axis cut K at O, the vertex. Draw the point F' so that O is the midpoint of FF'. Then we can draw the line through F' perpendicular to FF', which is the directrix.