Deriving an equation of a parabola I would like to understand the concept of deriving an equation, given values. E.g. Derive the equation of parabola whose vertex is at origin and focus $(-3,0)$.
From this, I reckon the ends are curving left with the focus at $(-3,0)$. I am aware the distance from the focus to the vertex is equal to distance from vertex to directrix.
I also think (not sure) that with this, the equation would be $$y^2 = 4px$$ $$y^2 = 4\times -3\times x$$
Can I conclude, \therefore that the equation is $y^2 = -12x$ ?
Thanks.
 A: As far as I know and by considering the coordinates of the focus $F(-3,0)$, the equation of parabola is: $$y^2=-2px$$ wherein $F(-p/2,0)$. So, here, $-p/2=-3$ and then $p=6$ and so $y^2=-12x$ would be the result.

A: If you mean by "derive" how the equation is obtained:
A parabola is the set of points "equidistant" from a line and a point external to that line, where "distance from a line" is taken to be the perpendicular distance.  So you want the points $ \ (x,y) \ $ for which  
$$ \sqrt{ ( x - [-3])^2  +  ( y - 0 )^2} \ = \ \sqrt{(x-3)^2} \ , $$
since the directrix is the vertical line $ \ x = 3 \ $ ; the left-hand side represents the distance from $ \ ( x,y ) \ $ to the focus $ \ (-3,0) \ $ and the right-hand side gives the perpendicular distance from $ \ ( x, y ) \ $ to the directrix, which would be measured along a horizontal line intersecting the directrix at $ \ (3, y) \ $ .
We can square both sides to obtain
$$ (x+3)^2 + y^2 \ = \ (x-3)^2 \ \Rightarrow \ x^2 + 6x + 9 + y^2 \ = \  x^2 - 6x + 9 \ $$
$$\Rightarrow \ 6x +  y^2 \ = \  - 6x  \ \Rightarrow \ y^2  =  -12x \ , $$
a "horizontal" parabola opening "to the left".  This argument can be generalized pretty easily to parabolas opening vertically as well.
The parabola is the "easy" conic section to derive; the ellipse and hyperbola require somewhat more effort...
A: I am assuming you are trying to find the parabola as the locus of points equidistant from focus and directrix.
The focus you are given at $F=(-3,0)$.
You are not given the directrix, but the vertex at $V=(0,0)$. Now the directrix is a line which is perpendicular to the line $FV$, on the opposite side of $V$ from $F$ and at the same distance from $V$ as $F$ is. [That's what the vertex is, plus the symmetry of the parabola]. So the directrix is the line $x=3$.
Now consider the point $(x,y)$ on the parabola. The distance from the line $x=3$ is $|(x-3)|$, and the distance from the point $(-3,0)$ is $\sqrt{(x+3)^2+y^2}$
Squaring the distances and equating the squares we find $(x-3)^2=(x+3)^2+y^2$ which simplifies to $y^2=-12x$.
The negative sign is because the directrix is to the right of the focus and reflects the orientation of the parabola.
