How do I find the equation of this parabola?

[Jarfi]

How do I find the equation of this parabola??

I am given a parabola with a "way line" l: x+y+2 = 0 and focus point F = (1;1).

How do I find it's equation?

I know I am suppost to shift it in some way and maybe mirror it, not sure how though. Need some help.

The equation of a parabole is y^2 = px, whereas x=-P/4 is the way line and F = (p/4;0) is the focus point.

I assume I am suppost to start with a parabole with focus point

F = (1;0), p = 4 and way line l: x=-1. Now I tried shifting it around (0;1) to get

F = (1;1), p = 4 and way line l: x=-2. I am almost there, but now I lack the Y, where on Earth does that Y come from? where do I go from there?

EDIT: I think going the opposite way is better but I still stop at the same spot:

I get l: x = -y -2 & F = (1;1).

1: shift around (0;-1) get:

x= -(y-1) - 2 = -y -1 & F = (1;0)

I am now closing in but all I need to do is get rid of the "-y". No idea how to do that.

EDIT:

Nevermind. I have found the solution, using the definition of a parabole lPFl = dist(p,l).

Case dismissed.

 

[Simon Bridge]

Well done.

One of the tricks is to work out what you have to do to orient the axis so the x-axis passes through the focus and the y-axis is parallel to the directrix ... then the inverse transform on the standard equation will get you the equation of the curve you want. In this case, that's just rotating the axis 45deg - which you see when you sketch it out.

 

[HallsofIvy]

I am given a parabola with a "way line" l: x+y+2 = 0 and focus point F = (1;1).

"Way line"? I am not familiar with that- I would say "directrix". In any case, the line of symmetry is perpendicuar to the directrix and goes through the focus. x+ y+ 2= 0 has slope -1 so the line of symmetry must have slope 1. A line with slope 1 is of the form y= x+ b. To go through (1, 1), we must have 1= 1+ b or b= 0. The line of symmetry is y= x.

Further, the vertex must be on that line of symmetry exactly half way between the focus and the directrix. y= x intersects x+ y+ 2= 0 when x+ x+ 2= 0 or x= -1, y= -1. The vertex is at ((-1+ 1)/2, (-1+ 1)/2)= (0, 0) (which simplifies the problem greatly- we do NOT need to translate the graph!).

You do not need to "shift" or "mirror", you need to rotate.

The distance from the vertex to the focus (the "focal length") is [itex]\sqrt{2}[/itex].
Write the equation of a parabola with vertex at (0, 0) and focal length [itex]\sqrt{2}[/itex].
Call the variables x' and y', reserving x and y for after the rotation.

Because we will have to rotate so that the y' axis becomes the line y= x and the x' axis become y= -x. That's rotation by [itex]\pi/4[/itex] radians so the cosine and sine are both [itex]\sqrt{2}/2[/itex].

How do I find it's equation?

I know I am suppost to shift it in some way and maybe mirror it, not sure how though. Need some help.

The equation of a parabole is y^2 = px, whereas x=-P/4 is the way line and F = (p/4;0) is the focus point.

I assume I am suppost to start with a parabole with focus point

F = (1;0), p = 4 and way line l: x=-1. Now I tried shifting it around (0;1) to get

F = (1;1), p = 4 and way line l: x=-2. I am almost there, but now I lack the Y, where on Earth does that Y come from? where do I go from there?

EDIT: I think going the opposite way is better but I still stop at the same spot:

I get l: x = -y -2 & F = (1;1).

1: shift around (0;-1) get:

x= -(y-1) - 2 = -y -1 & F = (1;0)

I am now closing in but all I need to do is get rid of the "-y". No idea how to do that.

EDIT:

Nevermind. I have found the solution, using the definition of a parabole lPFl = dist(p,l).

Case dismissed.