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!