Parabola equation with focus and directrix

[aisha]

Hi, I need to determine the equation of a parabola given the focus (2,3) and the directrix y=-1

I sketched out a parabola opening up wards with a vertex of (1,1)

I made two distance equations one for any point on the parabola to the focus, and one distance from the directrix to any point on the parabola.

I equated the two equations

sqrt((x+2)^2 + (y+3)^2)) = sqrt((y+1)^2))

I took the square root of both sides and then tried to expand and simplify
I got [tex] x^2 - \frac {1} {4}x + \frac {9} {8} = y [/tex]

there probably are mistakes since this is my first time doing this.

 

[HallsofIvy]

How did you get the vertex to be (1,1)??

A parabola is defined as the set of points such that the distance from each point (x,y) to the focus is the same as the distance from (x,y) to the directrix.
The distance from (1,1) to (2,3) is [tex]\sqrt{5}[/tex]. The distance from (1,1) to
y= -1 is 2. Those are not the same. (1,1) is not even ON the parabola.

Since, in this problem, the directrix is a horizontal line, the axis of the parabola is vertical. The vertex must be on the vertical line through the focus: x= 2. And, since the two distances must be the same, it must be exactly half way between the two:
y= (3+(-1))/2= 1. The vertex is at (2,3).
If you know that the standard formula of a parabola with vertex (a,b) and focus (a,c) is y= (1/4(b-c))(x-a)²+ b, you can just write down the formula.

I would consider it more interesting to do as you did- use the basic definition rather than memorizing a formula.

Let (x,y) be any point on the parabola. Then the distance from (x,y) to (2,3) is
[tex]\sqrt{(x-2)^2+ (y-3)^2}[/tex] (NOT (x+2) and (y+3)!). The distance from
(x,y) to the line y= -1 is [tex]\sqrt{(y+1)^2}= |y+1|[/tex]. The condition that (x,y)be on the parabola is [tex]\sqrt{(x-2)^2+ (y-3)^2}= |y+1|[/tex].
Squaring both sides (NOT taking the square root, as you wrote!)
(x-2)²+ (y-3)²= (y+1)²

x²- 4x+ 4+ y²- 6y+ 9= y²+ 2y+ 1

This is a parabola because the two "y²" terms cancel out.

x²- 4x+ 12= 8y so

y= (1/8)x²- (1/2)x+ 3/2

 

[Data]

The vertex is at (2,3).

You mean (2,1)