IA idea #6 - Directrix of a parabola

Consider the following property for parabolas.

Many students do not know what the directrix of a parabola is - which is a bit of a shame because then they are not aware of the fact that a parabola is the set of points (locus of points) that are equidistant from a line (directrix) and a point not on the line (focus).

Here is the outline of a way to start to explore (discover?) this property for parabolas.

Choose a fairly simple quadratic equation such that the vertex of the parabola is the origin or at least on the y-axis. Let's consider the graph of \(y = \dfrac{1}{4}{x^2}\). The derivative, \(\dfrac{{{\textrm{d}}y}}{{{\textrm{d}}x}} = \dfrac{1}{2}x\), gives the slope of the tangent at any point on the parabola. Select a point at which to find the equation of the tangent. Let's choose the point where \(x = 4\).  The coordinates of the point on the parabola are \(\left( {4,4} \right)\) and the slope of the tangent is \(\dfrac{{{\textrm{d}}y}}{{{\textrm{d}}x}} = \dfrac{1}{2}\left( 4 \right) = 2\). Thus, the equation of the tangent line at the point \(\left( {4,4} \right)\) is \(y = 2x - 4\). Now we need to find the point on the parabola where the tangent line will be perpendicular to the tangent line we already have, i.e. \(y = 2x - 4\).  Hence, need to find point where derivative is equal to \( - \dfrac{1}{2}\). This occurs at \(x = - 1\). The point of tangency is \(\left( { - 1,\dfrac{1}{4}} \right)\) and the equation of the tangent line is \(y = - \dfrac{1}{2}x - \dfrac{1}{4}\). The two perpendicular tangent lines, \(y = 2x - 4\) and \(y = - \dfrac{1}{2}x - \dfrac{1}{4}\), intersect at the point \(\left( {\dfrac{3}{2}, - 1} \right)\). These results are shown in the graph at left.

Let's repeat this procedure for another point on the parabola - choosing \(x = \dfrac{3}{2}\).  The equation of the tangent at \(\left( {\dfrac{3}{2},\dfrac{9}{{16}}} \right)\) is \(y = \dfrac{3}{4}x - \dfrac{9}{{16}}\). The point on the parabola where derivative is equal to \({ - \dfrac{4}{3}}\) is \(\left( { - \dfrac{8}{3},\dfrac{{16}}{9}} \right)\) - and the equation of the tangent at this point is \(y = - \dfrac{4}{3}x - \dfrac{{16}}{9}\).  The two perpendicular tangent lines intersect at \(\left( { - \dfrac{7}{{12}}, - 1} \right)\). The results for this 2nd set of perpendicular tangent lines are shown below right.  Note that the point of intersection of the two perpendicular tangent lines has a y-coordinate of -1; same as for the first pair above.

Since any two distinct points must be collinear, we must repeat the procedure for at least one more point on the parabola.

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Choosing a starting point of \(\left( { - 5,\dfrac{{25}}{4}} \right)\) and following the same procedure as before produces the two perpendicular tangent lines of \(y = - \dfrac{5}{2}x - \dfrac{{25}}{4}\) and \(y = \dfrac{2}{5}x - \dfrac{4}{{25}}\) that intersect at \(x = - \dfrac{{21}}{{10}}\).  Results for this 3rd set of perpendicular tangent lines is shown above.

So, if the parabola property I mentioned at the start of this blog entry was unknown then someone would hopefully conjecture the property from these three results.

This has potential for being a suitable topic for a student Exploration. I think a key aspect that would make it a strong Exploration is for a student to prove this property. This would require some thoughtful input from the student on setting up parameters and variables - and then organizing a clear algebraic proof annotated with clearly written text and supported by clearly labeled diagrams.  I think it is also conducive for a good student to think of some extensions to this property. For example, do any other conic sections (ellipse, hyperbola, etc) have a similar property?  Do other conic sections have a 'directrix'?  If so, what role does it play in constructing the particular conic section as a locus of points?  Is there a way to illustrate the property of parabolas shown above using another procedure?