The Parabola - The Burns Home Page

The General Equation of a Parabola with y-Axis as the Axis of SymmetryThe general equation of a parabola with the y-axis as the axis of symmetry and focus at(0, p) and directrix at y=-p isx24pyory14p2 x .The General Equation of a Parabola with x-Axis as the Axis of SymmetryThe general equation of a parabola with the x-axis as the axis of symmetry and focusat (p, 0) and directrix at x=-p isy24px x14p2 y .Note: The advantage ofx24pywrite the parabola as x a 24py b.format is the when given the vertex at (a, b), we canNote: The parabola opens up or to the right if p>0, down or to the left if p

ExampleThe general form of a parabola isand graph it.2y x y4 2 51 . Write the standard form of this conicSolution:Change into standard form by dividing by rearranging2y x y4 2 512y 2 y 514x2y y x2 1 51 4 12y1 4 x13This is in form y2 4px,which is a parabola thatopens to the rightThe parabola opens to the right, and its vertex is (-13,-1).

ExampleFind the equation for the locus of points that are equidistant from the point (0,2), andfrom the line y=-4. State the vertex and the focal length of the curve.Solution:Let P(x,y) be a point on the curve and D(x,-4) be a point on the directrix.The distance from the focus to P is the same as the distance from the D to P. That isPF=PD.PF PD2 2 22x 0 y 2 x x y 42 2 2 42x y y2 2 2x y y y yxx22 4 4 8 1612 y 12y12 1The vertex is (0,-1), and the focal length is 3 (12=4x3)

ExampleDetermine the directrix, the focus, and the equation of the parabola that passes throughthe point 1, 20, has vertex (0, 0), and has vertex on the x-axis. Sketch the graph ofthe parabola, and label its focus and directrix.Solution:Since the focus is one the x-axis, the equation has the form y2 41, 20is on the graph, we haveTherefore the focus is parabola is2y 20x.y24px2204 p120 4pp 5px. Now, since5,0and the directrix is the line x 5, and the equation of the

ExampleA radio dish at the VLA ( Very Large Array) at Socorro, New Mexico has the shape of aparabolic dish (the cross section through the centre of the dish is a parabola). This dish isabout 12 feet deep at the centre and has a diameter of 82 feet. How far from the vertexof the parabolic dish should the receiver be placed so that it is in the focus?Solution:We will draw a cross section of the dish with the parabola’s vertex on the x-axis. Thisgives us an equation of the form x2 4 py . Because the dish is 82 feet across, and 12feet deep, the point (41, 12) is on the parabola.Thereforex22 4py 41 4 p 12168148pp 35Thus the focus is (0, 35), or the receiver should be placed 35 feet from the vertex(bottom of the dish).