# Thread: Step by step parametrics?

1. ## Step by step parametrics?

Can someone run me through step by step on how to do find the equation of the locus of the point P(x,y) which moves so that it is always equidistant from the point S and the straight line L where S(1,2), L: y= -4
Can someone give me a step by step on how to find a and how to do the question? Thanks

2. Originally Posted by nerdzor
Can someone run me through step by step on how to do find the equation of the locus of the point P(x,y) which moves so that it is always equidistant from the point S and the straight line L where S(1,2), L: y= -4
Can someone give me a step by step on how to find a and how to do the question? Thanks
Distance of a point (a,b) from another point (m,n) is given by

$\displaystyle \sqrt{(a-m)^2 +(b-n)^2}$

Distance of a point (m,n) from a line
ax +by +c =0
is given by

$\displaystyle |\frac{(am+bn+c)}{\sqrt{a^2 +b^2}}|$

Let (h,k) be such a point

Thus its distance from (1,2) is equal to its distance from
y+4 =0

Hence

$\displaystyle \sqrt{(h-1)^2 +(k-2)^2} =\frac{(k+4)}{\sqrt{1^2}}$

This is the last equation that is to be simplified but be careful

.................................................. ..

3. Hello, nerdzor!

Find the equation of the locus of the point P(x,y) which moves so that
it is always equidistant from the point $\displaystyle S(1,2)$ and the line $\displaystyle L\!:\:y= -4$
First, make a sketch . . .
Code:
        |
|   *S(1,2)
|    \
|     \
----+------\---------
|       *P(x,y)
|       :
|       :
|       :
-4+ - - - * - - -  L
|       Q

The distance from $\displaystyle P\text{ to }S$ is: .$\displaystyle \sqrt{(x-1)^2 + (y-2)^2}$

The distance from $\displaystyle P$ to line $\displaystyle L$ is: .$\displaystyle PQ \:=\:y-(-4) \:=\:y+4$

Since $\displaystyle PS = PQ$, we have: .$\displaystyle \sqrt{(x-1)^2 + (y-2)^2} \:=\:y+4$

That is the equation of the locus, so technically, we're done.
. . But, of course, we are expected to simplify our equation.

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

. . Expand: .$\displaystyle x^2 - 2x + 1 + y^2 - 4y + 4 \:=\:y^2 + 8y + 16$

. . And we have: .$\displaystyle x^2 - 2x \:=\:12y + 11$

. . Add 1 to both sides: .$\displaystyle x^2 - 2x + 1 \:=\:12y + 12$

. . Factor: .$\displaystyle \boxed{(x-1)^2 \:=\:12(y+1)}$

The locus is a parabola.
. . It opens upward, and its vertex is at $\displaystyle (1,-1)$