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
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.
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 \(\displaystyle (x, \frac{1}{8}x^2+ 2)\).
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))}".
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.
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.