Arriving at parabola formula via distance formula

[ducmod]

Homework Statement

Hello!

Please, help me to get through equations. I can't derive the equation in the way suggested.

Here is the definition:

If we choose to place the vertex at an arbitrary point (h; k), we arrive at the following formula
re-deriving the formula from Denition 7.3. (If the vertex is at (0;0), then from the definition of parabola,
we know the distance from focus point (0; p) to a point (x; y) is the same as the distance from a point
on directrix (x;-p) to the same point (x; y)).
Using the Distance Formula we get:
The Standard Equation of a Vertical Parabola: The equation of a (vertical) parabola with vertex (h; k) and
focal length |p| is
(x - h)^2 = 4p(y - k)
If p > 0, the parabola opens upwards; if p < 0, it opens downwards.

So, if vertex is at (h, k) and there is a given point on a parabola at (x; y), then focus is at (h; k + p),
and point of a directrix is at (x; k - p).

Thus the distance formula should be (I am dropping the square root):
(x - h)^2 + (y - k)^2 = (y - k + p)^2

How did they come to (x - h)^2 = 4p(y - k) ?

Thank you!

Homework Equations

The Attempt at a Solution

 

[Simon Bridge]

Did you try subtracting (y-k)^2 from both sides, expanding the RHS, and grouping terms?

 

[ducmod]

Did you try subtracting (y-k)^2 from both sides, expanding the RHS, and grouping terms?

I did, but didn't arrive at the correct outcome. Your question implies that I have to give it another try )

 

[Simon Bridge]

Well it just means that I don't know what you've tried ... go over it more carefully - double check your starting point.
If P is a point on the parabola and D is the corresponding point on the directrix, then a parabola with focal point F has: |PF| = |PD|
From what you've written; P=(x,y), D=(x,k-p), F=(h,k+p) ...

 

[Samy_A]

Thus the distance formula should be (I am dropping the square root):
(x - h)^2 + (y - k)^2 = (y - k + p)^2

The LHS is not what you want.