Parabola, it’s scarily simple…

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):

parabola 1

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

parabola 3

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

parabola 2

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.

parabola 4

As point C is moved the parabola is traced out.

parabola 5

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





Filed under bisecting, conic sections, conics, construction, definitions, Uncategorized

2 responses to “Parabola, it’s scarily simple…

  1. 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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