Focus and directix for a parabola

A parabola is defined as the locus of point where its distance to a line (called directrix) and a point (called focus) are equal.

As an canonical example, we define the directrix to be $ x = -a $, and the focus be $ (a, 0) $, then the parabola will be defined as point such that

$ (x + a)^2 = (x - a)^2 + y^2 $

Simplifying we will get $ y^2 = 4ax $.

The interesting thing is to find directix and focus from an arbitrary (let's say, still axis parallel) parabolas. It is best illustrated with an example

Suppose the parabola is $ -x^2 - 2y + 2x = 12 $

Now we have $ -x^2 $, so the parabola open downwards, it is best to deal with the positive coefficients, so we do:

$ -12 - 2y = x^2 - 2x $

Next we complete the square, this is trivial

$ -11 - 2y = (x - 1)^2 $

Last we just fit it into the form we want:

$ 4\frac{1}{2}(y + \frac{11}{2}) = -(x - 1)^2 $

So we know it is just  a shift of the standard form $ 4ay = x^2 $, the vertex is $ (1, \frac{-11}{2}) $, the focus is $ (1, -6) $, the directrix is $ y = -5 $. The focus and directix are found by moving $ a = \frac{1}{2} $ units from the vertex.

Here is a plot showing the elements

Last but not least, it is always nice to check answer:

$ (y + 5)^2 = (x - 1)^2 + (y + 6)^2 $

$ y^2 + 10y + 25 = x^2 - 2x + 1 + y^2 + 12y + 36 $

$  -x^2 - 2y + 2x = 12 $

So yes, we have got the right answer!