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Equation of Parabola with Given Vertex, Point

Date: 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,

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!   
Associated Topics:
High School Calculus
High School Equations, Graphs, Translations

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