My Math Forum  

Go Back   My Math Forum > High School Math Forum > Geometry

Geometry Geometry Math Forum


Thanks Tree2Thanks
  • 1 Post By skipjack
  • 1 Post By v8archie
Reply
 
LinkBack Thread Tools Display Modes
July 25th, 2014, 01:24 AM   #1
Newbie
 
Joined: May 2014
From: Sydney

Posts: 16
Thanks: 0

Exclamation 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?
Monokuro is offline  
 
July 25th, 2014, 02:00 AM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 16,605
Thanks: 1201

The relevant equation is x² + (y - 4)² = (y + 4)², i.e., y = x²/16. The locus is a parabola.
Thanks from ZardoZ

Last edited by skipjack; August 3rd, 2014 at 06:24 PM.
skipjack is offline  
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?
Monokuro is offline  
July 25th, 2014, 06:03 AM   #4
Math Team
 
Joined: Nov 2010
From: Greece, Thessaloniki

Posts: 1,989
Thanks: 131

Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus
Quote:
Originally Posted by Monokuro View Post
Hello, thanks I know it's a parabola, but how do you find point P?
??? Skipjack found the locus of P.
ZardoZ is offline  
August 2nd, 2014, 06:40 PM   #5
Senior Member
 
Mr Davis 97's Avatar
 
Joined: Feb 2014
From: Louisiana

Posts: 156
Thanks: 6

Math Focus: algebra and the calculus
Quote:
Originally Posted by Monokuro View Post
Hello, thanks I know it's a parabola, but how do you find point P?
Skipjack simply used the distance formula. Also, point P is not the solution. The locus of point P is the sought after answer.
Mr Davis 97 is offline  
August 2nd, 2014, 07:37 PM   #6
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 6,451
Thanks: 2125

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by Monokuro View Post
Hello, thanks I know it's a parabola, but how do you find point P?
The distance, $D$, of point $A(a_1,a_2)$ from point $B(b_1,b_2)$ is given (as MrDavis said) by the distance formula
$$D^2 = (a_1-b_1)^2 + (a_2-b_2)$$
(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
$$D_p^2 = (x - 0)^2 + (y - 4)^2 = x^2 + (y-4)^2$$

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 $(x,-4)$
$$D_l^2 = (x-x)^2 + (y + 4)^2 = (y + 4)^2$$

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 + (y-4)^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*}
Thanks from ZardoZ
v8archie is online now  
Reply

  My Math Forum > High School Math Forum > Geometry

Tags
locus



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Known locus? Mathmax Algebra 0 May 22nd, 2011 12:09 PM
Locus of centroid roark Algebra 0 November 26th, 2010 08:43 AM
Locus julian21 Algebra 1 November 8th, 2010 08:26 PM
Need help with locus Goldenglove Complex Analysis 0 December 15th, 2009 11:12 AM
LOCUS symmetry Algebra 6 January 30th, 2007 03:48 AM





Copyright © 2017 My Math Forum. All rights reserved.