Parabola 6Given the axis and two points (P1 and P2) on the curve, construct a parabola.  Here I have moved the axis into a horizontal position to make this construction easier to follow in terms of abscissa and ordinate. Most of the following construction can be gleaned from the diagram, but here are the conditions. Through the given points construct lines parallel and perpendicular to the given axis. Here P1A and P2B are parallel to the axis, while P1B is perpendicular to the axis, intersecting it at G, and P2A is perpendicular to the axis, intersecting it at E. Let AB intersect the axis at C. Let P1C intersect P2A at D. Draw BD, and construct a line through E parallel to BD, intersecting BP2 at H. Through H construct a line perpendicular to the axis, meeting it at V, the axial vertex.  Now to prove that V is actually the vertex, let P1 and P2 have abscissas x1 and x2, and ordinates y1 and y2, where the abscissa is measured left along the axis from V and the ordinate is measured perpendicular to the axis on either side. This condition must be satisfied:  Use the parallel lines and similar triangles to follow this:  So V is the vertex. To complete the construction see Parabola 3, where a parabola is constructed given the axis, axial vertex, and a point on the curve. |