Find the equation of the parabola whose
focus is (2,3) and directrix is x=12

Let's plot the focus (2,3), marked with a <



Now let's draw the directrix, which is a
vertical line whose equation is x=12, 12
units right of the y-axis.



We can now tell that the parabola opens to 
the left, and has equation of the form

(y - k)² = 4p(x - h)

where (h, k) is the vertex, and p = the distance
between the focus and the vertex, which is also
the distance from the vertex to the directrix.
However this distance is taken as NEGATIVE since
the parabola must open to the left.

The vertex is the point which is half-way between the
focus and the directrix. Since the distance from the 
focus to the directrix is 9 units to the right, the 
vertex is located half of 9 units, or 4.5 units to 
the right of the focus. That point, which is 4.5 units 
to the right of the focus (2,3), is the point 
(2+4.5, 3) or (6.5, 3). So that's the vertex
so (h, k) = (6.5,3)

Now we plot the vertex with the mark =  



The only thing left to find is p, which is -4.5. It
has absolute value 4.5 because that is the distance 
from the focus to the vertex, and also the distance
from the vertex to the directrix.  And since the
parabola opens left, we make it negative, so
p = -4.6

So now we substitute (h, k) = (6.5, 3) and p = -4.5

into

(y - k)² = 4p(x - h)

and get

(y - 3)² = 4(-4.5)(x - 6.5)

(y - 3)² = -6(x - 6.5)

and the graph is

 

Edwin