# find the locus of a moving point P in algebraic form which is equidistant from the point (0,4) and the x-axis

#### chunwai

##### New member
Hello,

Please explain what is the question mean and how this answer get.
find the locus of a moving point P in algebraic form which is equidistant from the point (0,4) and the x-axis

#### Harry_the_cat

##### Senior Member
Draw a diagram and mark in the point (0, 4).

Now dot in some points which are the same distance from (0, 4) as they are from the x-axis (note that the distance between a point and a line is measured perpendicular to the line.

Call any one of these point (x, y).

Write an expression which represents the distance between (x, y) and (0, 4).
Write another expressions which represents the distance between (x, y) and the x-axis.
Equate these wo expressions and rearrange to get y = …..

Give it a go.

BTW the answer will be an equation involving x and y NOT the expression you have stated.

#### HallsofIvy

##### Elite Member
Let (x, y) be such a point. The distance to (0, 4) is $$\displaystyle \sqrt{x^2+ (y-4)^2}$$ and the distance to the x-axis is y. To be "equidistant" we must have $$\displaystyle y= \sqrt{x^2+ (y- 4)^2}$$. Start by squaring both sides: $$\displaystyle y^2= x^2+ (y- 4)^2= x^2+ y^2- 8y+ 16$$. The two "$$\displaystyle y^2$$" terms cancel to get $$\displaystyle x^2- 8y+ 16= 0$$, $$\displaystyle 8y= x^2+ 16$$ so $$\displaystyle y= \frac{1}{8}x^2+ 2$$.

Since the question asked about a "point" the answer is $$\displaystyle (x, \frac{1}{8}x^2+ 2)$$.

#### Harry_the_cat

##### Senior Member
Since the question asked about a "point" the answer is $$\displaystyle (x, \frac{1}{8}x^2+ 2)$$.
I always thought the locus of a point was given as an equation.

#### HallsofIvy

##### Elite Member
A locus is a set of points. Of course, the points can be given as "$$\displaystyle y= \frac{1}{8}x^2+ 2$$" with the understanding that the points are "(x, y)". But "y= f(x)" is a shortcut for "{(x, f(x))}".

#### Dr.Peterson

##### Elite Member
So, the answer is really $$\displaystyle \{(x, \frac{1}{8}x^2+ 2): x\in \mathbb{R}\}$$. It's a set of points, not a point.

But, yes, in my experience, the answer is usually given as an equation, meaning "the set of points that satisfy ...". That is, the locus is the graph of the equation.