# Locus question

• Jul 19th 2008, 03:10 AM
withyou
Locus question
Hi,
I have this problem:

If 'R' is the fixed point (3,2) and 'P' is the movable point (x,y)
Find the equation of the locus of 'P' if distance PR is twice the distance from P to y=-1

Thanks!
• Jul 19th 2008, 05:17 AM
mr fantastic
Quote:

Originally Posted by withyou
Hi,
I have this problem:

If 'R' is the fixed point (3,2) and 'P' is the movable point (x,y)
Find the equation of the locus of 'P' if distance PR is twice the distance from P to y=-1

Thanks!

$\displaystyle PR=\sqrt{(x-3)^2 + (y-2)^2}$.

Distance of P from the line y = -1 is $\displaystyle y + 1$.

Therefore ......
• Jul 19th 2008, 05:30 AM
Soroban
Hello, withyou!

Did you make a sketch?

Quote:

$\displaystyle R$ is the fixed point (3,2), and $\displaystyle P$ is the movable point (x,y)

Find the equation of the locus of $\displaystyle P$
if distance $\displaystyle PR$ is twice the distance from $\displaystyle P$ to $\displaystyle y=-1$

Code:

        |         |                P         |                o (x,y)         |              *  :         |          *    :         |        *        :         |  R o          :         |  (3,2)        :     ----+-----------------:------         |                :Q       - + - - - - - - - - o - -         |              (x,-1)         |

We have: .$\displaystyle PR \:=\:2\!\cdot\!PQ \quad\Rightarrow\quad \sqrt{(x-3)^2+(y-2)^2} \;=\;2(y+1)$

Square both sides: .$\displaystyle (x-3)^2 + (y-2)^2 \;=\;4(y-1)^2$

Expand: .$\displaystyle x^2-6x + 9 + y^2 - 4y + 4 \;=\;4y^2 + 8y + 4$

. . . . . . .$\displaystyle x^2 - 6x - 3y^3 - 12y \;=\;-9$

Complete the square:

. . $\displaystyle (x^2 - 6x \qquad) - 3(y^2 + 4y \qquad) \;=\;-9$

. . $\displaystyle (x^2 - 6x \:{\color{blue}+\:9}) - 3(y^2 + 4y \:{\color{red}+\:4}) \;=\;-9 \:{\color{blue}+\:9} \: {\color{red}-\:12}$

. . . . . . . . $\displaystyle (x-3)^2 - 3(y+2)^2 \;=\;-12$

Divide by -12: .$\displaystyle \boxed{\;\frac{(y+2)^2}{4} - \frac{(x-3)^2}{12} \;=\;1\;}$