No distances, no circles, and you can easily derive an equation.

Just a right angled triangle.

First, the definition of a parabola from the focus and directrix.

Pick a line, the directrix, and a point (B) not on that line (the focus):

Find the line at right angles, passing through a point (C) on that line.

Now find the line from B to C, and the midpoint of BC, which will be D.

Find the line at right angles to BC from D, and the intersection of this line and the vertical line, E, is a point on the parabola.

As point C is moved the parabola is traced out.

The picture is completed with the line BE. Check it!

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I love that approach! I’ve always taught it using the definition, but putting the vertex at the origin and, for a vertical parabola opening upwards, I would place the vertex on the origin and place the focus on (0,p) and directrix at

y = -p. Then I could place 2 points on the parabola: (2p,p) and (-2p, p) this would give me 3 definite points on the parabola. I would sketch the parabola through the 3 points. I was limited to showing only + and – parabolas that we’re symmetric to the x or y axis.

I look forward to playing around with your approach! I haven’t posted anything on my website for years!

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