Question 198607
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It actually does make a perverse kind of sense.  What you have is a degenerate parabola, actually a straight line that isn't a parabola at all.  Here's why.


Since the given directrix is a vertical line, the axis of symmetry of this parabola is horizontal.  The vertex form of the equation of a parabola with vertex at (<i>h</i>,<i>k</i>) and directrix <i>x</i> = <i>h</i> - <i>p</i> is:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (y - k)^2 = 4p(x - h)]


So using the equation of the directrix to calculate <i>p</i> when (<i>h</i>,<i>k</i>) = (3, 0) and <i>x</i> = 3:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = h - p = 3 - p = 3 \ \ \Rightarrow\ \ p = 0]


So the equation becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (y - 0)^2 = 4(0)(x - 3) = 0 \ \ \Rightarrow\ \ y = 0]


Which is the equation of the <i>x</i>-axis.


Geometrically speaking, a parabola is formed by the intersection of a plane and a cone where the plane is parallel to one of the generators of the cone.  But if you move the plane to a point where the generator of the cone actually lies <i>wholly in</i> the plane, the parabola degenerates to a straight line.  The distance from the focus to the vertex (and the distance from the directrix to the vertex) is a function of the distance from the axis of symmetry that lies in the plane and the generator of the cone.  When that distance is zero, you have no parabola.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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