Parabola 9Given one diameter (hence, the direction of all diameters) and three points (P1, P2, P3) on the curve, construct the parabola.  I am rather pleased with this one because it is all my own. The construction itself is not so complicated, but the proof may be. For that reason the proof of the construction is set aside in the separate page Parabola 9 Support. The Construction Construct line k through point P1 and perpendicular to the given diameter. Construct diameters through all three given points. Let the diameters through P2 and P3 meet k at points A and B, respectively. Dilate those two points with respect to P1, using scale factor 1/2. Through Aʹ construct a line perpendicular to P1P2, and through Bʹ construct a line perpendicular to P1P3. These lines meet at C, which lies on the axis of the parabola. Construct the axis through C, parallel to the given diameter. The perpendicular bisector of P1C meets the axis at F, the focus.  Construct the circle with center P1, passing through F. The diameter through P1 meets this circle on the directrix. On this part some style points may be earned. That diameter and circle actually meet at two points, only one of which is on the directrix. The correct choice is plain to see here, as the positions of the given points show that the parabola must have concavity upward and to the right, meaning the directrix must be downward and left from the focus. But to pass the drag test the construction should take care of itself while the points are shifted around. Some imagination is required here. The Parabola can now be constructed from the focus and directrix. |