July 25th, 2014, 01:24 AM  #1 
Newbie Joined: May 2014 From: Sydney Posts: 16 Thanks: 0  Locus
Question: Find the locus of a point P(x,y) such that its distance from the point (0,4) is equal to its distance from the line y=4. How do you solve this? 
July 25th, 2014, 02:00 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 17,740 Thanks: 1361 
The relevant equation is xÂ² + (y  4)Â² = (y + 4)Â², i.e., y = xÂ²/16. The locus is a parabola.
Last edited by skipjack; August 3rd, 2014 at 06:24 PM. 
July 25th, 2014, 02:16 AM  #3 
Newbie Joined: May 2014 From: Sydney Posts: 16 Thanks: 0 
Hello, thanks I know it's a parabola, but how do you find point P?

July 25th, 2014, 06:03 AM  #4 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 132 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  
August 2nd, 2014, 06:40 PM  #5 
Senior Member Joined: Feb 2014 From: Louisiana Posts: 156 Thanks: 6 Math Focus: algebra and the calculus  
August 2nd, 2014, 07:37 PM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,878 Thanks: 2240 Math Focus: Mainly analysis and algebra  Quote:
$$D^2 = (a_1b_1)^2 + (a_2b_2)$$ (Note that I have the distance squared in that line, because that is how we'll use it). This directly leads us to find that the distance, $D_p$, of the point $P(x,y)$ from $(0,4)$ is $$D_p^2 = (x  0)^2 + (y  4)^2 = x^2 + (y4)^2$$ The distance of the point $P(x,y)$ from the line $y = 4$ is the perpendicular distance between the two. That is, the distance, $D_l$, of the point $P(x,y)$ from the the point $(x,4)$ $$D_l^2 = (xx)^2 + (y + 4)^2 = (y + 4)^2$$ The question asks that these distances be equal, giving us Skipjack's solution: \begin{align*} D_p &= D_l \\ D_p^2 &= D_l^2 \\ x^2 + (y4)^2 &= (y + 4)^2 \\ x^2 + \cancel{y^2} 8y + \cancel{16} &= \cancel{y^2} + 8y + \cancel{16} \\ x^2 &= 16y \\[12pt] y &= \tfrac{1}{16}x^2 \\ \end{align*}  

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