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 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: 16,377 Thanks: 1174 The relevant equation is xÂ² + (y - 4)Â² = (y + 4)Â², i.e., y = xÂ²/16. The locus is a parabola. Thanks from ZardoZ 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
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Quote:
 Originally Posted by Monokuro Hello, thanks I know it's a parabola, but how do you find point P?
??? Skipjack found the locus of P.

August 2nd, 2014, 06:40 PM   #5
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Quote:
 Originally Posted by Monokuro 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.

August 2nd, 2014, 07:37 PM   #6
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Quote:
 Originally Posted by Monokuro 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*}

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