Equation of Parabola with Given Vertex, PointDate: 8/4/96 at 10:19:38 From: Anonymous Subject: Equation of Parabola with Given Vertex, Point I have a problem I cannot solve except by trial and error. "Find an equation of all parabolas containing the origin whose vertex is at the point (2,1)." By trial and error I arrived at x=-2(y-1)^2+2; y=-1/4(x-2)^2+1. Also, how do you find the directrix? Thanks again, Everette Date: 8/4/96 at 14:0:53 From: Doctor Robert Subject: Re: Equation of Parabola with Given Vertex, Point The general form of a parabola is y-k = a(x-h)^2 where (h,k) is the vertex and a is a constant. Now, if this parabola is to also contain the origin (0,0), then -k = ah^2. Or a = -k/h^2. So your equation for the parabola opening in the vertical direction is correct. To get the other equation you would start out with x-h = a(y-k)^2 and use the same logic. You also have the correct equation here. These are the only two parabolas that can have a vertex at (2,1) and contain the origin. If (2,1) were just a point, rather than the vertex, there would be many more possibilities. I hope that this helps. -Doctor Robert, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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