First, we need to determine whether this is a horizontal or a vertical parabola. Because we go left  units to get from the vertex to the focus, we know that this is a horizontal parabola with .

Recall the standard equation of a horizontal parabola:

, where  is the vertex and  is the focal length.

Since the question gives us the vertex, we know that  and .

Now, plug all the given information to get the standard equation of a horizontal parabola:

First, we need to determine whether this is a horizontal or a vertical parabola. Because we go left  units to get from the focus to the directerix, we know that this is a horizontal parabola.

Recall that the distance from the focus to the directerix can be given as . .

Recall the standard equation of a horizontal parabola:

, where  is the vertex and  is the focal length.

Next, we need to find the vertex. Since this is a horizontal parabola, the -coordinates of the focus and the vertex will be the same. To find the -coordinate, go left  units from the focus. The vertex is then located at . Then,  and .

Since the focus is located to the right of the vertex, we know that .

Now that we have all the parts needed to write the standard equation for this parabola:

Recall the standard equation of a horizontal parabola:

, where  is the vertex and  is the focal length.

Start by isolating the  terms.

Complete the square on the left. Make sure to add the same amount to both sides of the equation!

Factor both sides of the equation to get the standard form of a horizontal parabola.

Since the directrix is to the right of the focus, and the vertex should be some distance in between, we know that this parabola opens to the left, and is then negative, in the form:

where p is the distance from the vertex to the focus/directrix, and the vertex is .

Since the directrix is at and the focus has an x-coordinate of 1, the distance from the focus to the directrix is . The vertex is in the middle, or 3 away from each, so . The vertex will be at .

Our equation then is:

or

This equation has a directrix is at , and the y-coordinate of the focus is 1. This means that the distance from the focus to the directrix is 4. The vertex will be halfway between, so 2 away from each. That means that p is 2. If the focus is , then the vertex is . 

Since the directrix is a horizontal line, this equation will be in the form , with vertex .

Since the focus is below the directrix, the parabola opens down, and the equation will be negative.

Our equation is: or 

To be in standard form, the equation for a parabola must be written in one of the following ways:

     OR     

THe problem given has the square around the x term, so it's going to end up loking like the standard form on the left.

First, we square the right side

Lastly, we need the y by itself, so we add 3 to both sides

The vertex form for a parabola is given below:

The vertex coordinates are  or  as given in the problem statement.

Now solve for a using the point it crosses , by plugging the point into the equation.

Plug a back to find the final answer:

The vertex form for a parabola is given below:

The vertex coordinates are  or  as given in the problem statement.

Now solve for a using the point it crosses , by plugging the point into the equation.

Plug a back to find the final answer:

The vertex form for a parabola is given below:

The vertex coordinates are  or  as given in the problem statement.

Now solve for a using the point it crosses , by plugging the point into the equation.

Plug a back to find the final answer:

To find the focus from the equation of a parabola, first set the equation to resemble the form  where  represents any numerical value.

For our problem, it is already in this form.

Therefore, 

.

Solve for  then .

The focus for this parabola is given by .

So,  is the focus of the parabola.

The directrix is represented as .

Therefore, the directrix for this problem is .