1. ## parabola

Find an equation that says that P=(x,y) is equidistant from
F= (2.0) and the y-axis. Plot four points that fit this equation.
The configuration of all such points P is called a parabola.

How come the equation (x-2)squared + y squared = xsquared + (y-y1)squared doesn't work?
This does not seem to be a parabola.

2. You need to know one thing, that the shortest distance between a point and a line is the distance that is perpendicular to the line.

Since $\displaystyle \displaystyle (2, 0)$ is on the $\displaystyle \displaystyle x$ axis, which is already perpendicular to the $\displaystyle \displaystyle y$ axis, then the shortest distance from this point to the $\displaystyle \displaystyle y$ axis is the distance along the $\displaystyle \displaystyle x$ axis, i.e. $\displaystyle \displaystyle 2$ units.

So that means that the distance between $\displaystyle \displaystyle (x, y)$ and $\displaystyle \displaystyle (2, 0)$ must also be $\displaystyle \displaystyle 2$ units.

Therefore $\displaystyle \displaystyle \sqrt{(x - 2)^2 + (y - 0)^2} = 2$.

Simplify this for the equation of your parabola.

3. I get (x-2)squared + ysquared = 4
Isn't this a circle?

4. Originally Posted by Veronica1999
Find an equation that says that P=(x,y) is equidistant from
F= (2.0) and the y-axis. Plot four points that fit this equation.
The configuration of all such points P is called a parabola.

How come the equation (x-2)squared + y squared = xsquared + (y-y1)squared doesn't work?
This does not seem to be a parabola.
Let $\displaystyle (x,y)$ be a point on the parabola. The distance to the focus is

$\displaystyle d^2=(x-2)^2+(y-0)^2$

Now the distance from the point (x,y) to the directrix is always measured perpendicular to the directrix. So in our case it is just the x coordinate. so

$\displaystyle d^2=x^2$ equating the two gives

$\displaystyle x^2=(x-2)^2+y^2/ \iff x^2=x^2-4x+4+y^2 \iff y^2=4x-4=4(x-1)$

### find the relation between x & y such that point p(x,y) is equidistant fron the point A ( -5,3) and (7,2)

Click on a term to search for related topics.