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Purdue MA 15400 - Parabolas

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MA 154 Lesson 25 DelworthMA 154 Lesson 25 DelworthSection 11.1 ParabolasA parabola is the set of all points in a plane equidistant from a fixed point F (the focus) and a fixed line l (the directrix) that lie in the plane.The axis of the parabola is the line through F that is perpendicular to the directrix.The vertex of the parabola is the point V on the axis halfway from F to l. The vertex is the point on the parabola that is closest to the directrix.The vertex and focus are points; the directrix and axis are lines.The distance from the vertex to the focus and from the vertex to the directrix is p units.The distance from the focus to the directrix is 2p units.If the parabola has a vertical axis and its vertex is at (0, 0), its formula is: x24 pyIf the parabola has a horizontal axis and its vertex is at (0, 0), its formula is: y24 pxThe proof of this is not too bad. I will do it for a vertical parabola:Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix.x28y2y2 5x1MA 154 Lesson 25 DelworthSection 11.1 ParabolasIf we take the standard equation of a parabola and replace x with x – h and y with y – k, then x24 pybecomes x  h 24 p y  k  and y24 px becomes y  k 24 p x  h  with vertex V(h, k).Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix.y  2 212(x  1)x  3 213y  2 y214y  4x  45 0x2 20y 10Sketch the parabola described and find an equation for the parabola.V(3, –1), F(3, 2) V(–2, 3), F(–6,


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Purdue MA 15400 - Parabolas

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