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Recently, I learnt that other than using a quadratic equation to describe a parabola, it is also possible to say that a parabola is the locus of points that is equidistant from a focus and a directrix.

The idea of the parabola having points equidistant from a single focus and a line (directrix) seems like an acceptable idea, but for some reason,

I've been wondering how we know that this definition works? Is there a way to show that all the points are equidistant from the focus and directrix?

helpme
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  • Related (duplicate?): "General Equation for Parabola - Proof". The question statement doesn't say so (a comment does), but the asker wants specifically to use the focus-directrix definition of the parabola to get to the quadratic equation. – Blue Sep 03 '19 at 07:52
  • This was the original definition of the Ancient Greeks, before coordinates were introduced, just as an ellipse was the locus of points such that the sum of their distances to the two foci was constant (and a similar definition for the hyperbola) – Bernard Sep 03 '19 at 08:07
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    @Bernard I believe the original definition was rather based on conic sections. – Jean-Claude Arbaut Sep 03 '19 at 08:12
  • @Jean-ClaudeArbaut:: You mean Menæchmus? I wrote my comment remembering what I was taught while high school at the end – quite some time ago, so my memories may be not absolutely exact… ;-) – Bernard Sep 03 '19 at 11:35
  • @Bernard According to Lebesgue's Les coniques, conics were first studied by Menaechmus and later Apollonius, and both defined conics by intersecting a cone by a plane. There is a much more detailed account in Heath's History of Greek Mathematics. – Jean-Claude Arbaut Sep 03 '19 at 11:49

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First, make a change or coordinates (translation and rotation) to get an equation $y=ax^2$ for your parabola. Take an unknown focus $F$ as $(0,b)$ and a directrix as $y=-b$: it must be $-b$ because for $x=0$, the point $(0,0)$ on the parabola is equidistant from $(0,b)$ and the directrix. Let $M(x,ax^2)$ be a moving point on the parabola, and $H(x,-b)$ the orthogonal projection on the directrix.

Now, let's check the distances, or equivalently the squares of the distances:

$$MH^2=(ax^2+b)^2=a^2x^4+2abx^2+b^2$$ $$MF^2=x^2+(ax^2-b)^2=a^2x^4+(1-2ab)x^2+b^2$$

They are equal if $2ab=1-2ab$, that is $b=\dfrac{1}{4a}$.

Jean-Claude Arbaut
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