Ask Me Help Desk - A parabola with directrix at x=3 and focus at f{1,-1},find its equation

a parabola with directrix at x=3 and focus at f{1,-1},find its equation

i want to know the steps to follow so that i can know how to solve simillar questions.what is the value of p,how do i know that.

the squation is (y+1)^2=4(x-2) and p=1,that is according to my understanding.

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The directrix and the focus are the same distance from the vertex.

So, the parabola opens to the left and has vertex at (2,-1).

Therefore, it has the form

$(y-k)^{2}=-4p(x-h)$

Where p is the distance from the focus to the vertex or from the vertex to the directrix. Thus, p=1, h=2 and k=-1.

So, its equation is

$(y+1)^{2}=-4(x-2)$

$x=\frac{-1}{4}y^{2}-\frac{1}{2}y+\frac{7}{4}$

First thing is to recognize that if the directrix is defined by x = 3, this means the parabola is "lieing on its side" and the standard formula is of the form:
$ (y-y_1)^2 = 4p (x-x_1) $

where $(x_1, y_1)$ is the vertex and $p$ is the distance from vertex to directrix, (which also happens to be the distance from the vertex to the focus of the parabola). So now you need to determine the coordinates of the vertex. Because the vertex is midway between the directrix line (x = 3) and focus (1,-1), it is at $(\frac {3+1} 2 , -1)$ , or (2,-1). These are the coordinates for $(x_1, y_1)$ . The distance between the vertex and the directrix is then 2-3 = -1. Note that this is a negative value, because the vertex is to the left of the directrix, which means that the parabola opens up to the left. So the value to use for $p$ is -1. Hence the formula is:

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

Note that if you had been told that the directrix was y = something, that would mean that the parabola is of the form $(x-x_1)^2 = 4p (y-y_1)$ , with $p$ positive if the focus is above the directrix, and $p$ negative if the focus is below the directrix.