Finding Equations of ParabolasDate: 12/12/96 at 19:02:20 From: Kris Duguay Subject: Conics (parabolas) The problem that I am pondering and trying to solve deals with the special conic, the parabola. I have drawn the graph and plotted the given information and I just cannot see where to begin. I would dearly appreciate it if you could help me with the following: Determine the equation of the parabola whose vertex is at (1,3) and whose directrix is y=x. Date: 12/12/96 at 21:56:00 From: Doctor Pete Subject: Re: Conics (parabolas) The directrix is related to the focus by the fact that the parabola is the locus of points for which the distance to the focus and perpendicular distance to the directrix is equal. There are a few ways to go about solving this problem, but I think the easiest would be to introduce a transformed coordinate system in which the directrix is parallel to an axis; that is, let's introduce a rotated coordinate system (x', y'), in which the x'-axis is Pi/4 degrees clockwise of the x-axis. To do this, recall that a transformed coordinate axis which is rotated about the origin through an angle T is given by: [ x' ] [ Cos T Sin T ][ x ] [ x Cos T + y Sin T ] | | = | || | = | | [ y' ] [ -Sin T Cos T ][ y ] [ -x Sin T + y Cos T ] So if T = Pi/4, Cos T = Sin T = 1/Sqrt[2], and this becomes: x' = (x + y)/Sqrt[2] y' = (y - x)/Sqrt[2] Note that under this transformation, the line y = x becomes the line y' = 0, and we see that the vertex becomes: (1,3) --> ((1+3)/Sqrt[2]', (3-1)/Sqrt[2]') = (2*Sqrt[2]', Sqrt[2]') Now, how do we find the equation of the parabola under this transformed coordinate system? It's easy because we now know that the distance from the vertex to the directrix is simply Sqrt[2]'. Thus the focus must be located at: F' = (2*Sqrt[2]', 2*Sqrt[2]') since the vertex is halfway between the directrix and focus. Now, we use the focus-directrix definition of the parabola to find its equation. If some point (x', y') is on the parabola, then its perpendicular distance to the directrix is simply y'. This must then be equal to its distance to the focus, so: Sqrt[(x' - 2*Sqrt[2])^2 + (y' - 2*Sqrt[2])^2] = y' Simplifying, we obtain: (x' - 2*Sqrt[2])^2 + (y' - 2*Sqrt[2])^2 = (y')^2 (x')^2 - 4*Sqrt[2] x' + 8 + (y')^2 - 4*Sqrt[2] y' + 8 = (y')^2 (x')^2 - 4*Sqrt[2] x' + 16 = 4*Sqrt[2] y' y' = (x')^2/(4*Sqrt[2]) - x' + 4/Sqrt[2] = (x')^2/(4*Sqrt[2]) - x' + 2*Sqrt[2] This is our equation for the parabola in the transformed coordinate system. Thus, to get the equation of the parabola in the original coordinate system, we apply the transformations to the equation. Wherever we see an x', we replace it by (x+y)/Sqrt[2], and wherever we see a y', we replace it by (y-x)/Sqrt[2]. This gives us: (y-x) ((x+y)/Sqrt[2])^2 (x+y) ------- = ----------------- - ------- + 2*Sqrt[2] Sqrt[2] 4*Sqrt[2] Sqrt[2] (x+y)^2 (y-x) = ------- - (x+y) + 4 2*4 8(y-x) = x^2 + 2xy + y^2 - 8(x+y) + 32 x^2 + 2xy + y^2 - 16y + 32 = 0 This is the desired result. If we substitute the point (1,3) into this equation, it indeed returns 0. Note that I didn't graph a thing - the entire solution is algebraic, although it often helps to picture what's going on, especially when one uses rotation matrices. While the solution I have given seems quite long, it isn't all that much computation compared to trying to do it without rotating the coordinate axes. The key step is to understand the focus-directrix relationship; the idea of the directrix has a direct geometric interpretation, which generalizes to any conic, not just the parabola. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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