How to find the coefficients of a parabola if only its focus and directrix are known

This is related to my previous question Explanation of method for finding the intersection of two parabolas.

I am trying to understand the math behind a piece of code that calculates the coefficients of a parabola ($a$, $b$ and $c$), knowning only its focus and directrix.

The relevant part of the code (in C++) is:

double dp = 2.0 * (p->y - y);
double a1 = 1.0 / dp;
double b1 = -2.0 * p->x / dp;
double c1 = y + dp / 4 + p->x * p->x / dp;


I know that p->x and p->y are the x and y of the focus and y is the y of the directrix.

And a1, b1 and c1 are the calculated coefficients of the standard form of the parabola equation.

The code is used only in a special case, where the directrix is always parallel to the x axis and the focus is always above the directrix.

My attempt at reconstructing the formulas from the code is:

$a = \frac{1}{2( y_{f} - y_{d})}$

$b = \frac{-2x_{f}}{2( y_{f} - y_{d})}$

$c = y_{d} + \frac{2( y_{f} - y_{d})}{4} + \frac{x_{f}^2}{2( y_{f} - y_{d})}$

where $x_{f}$ and $y_{f}$ are the $x$ and $y$ of the focus and $y_{d}$ is the $y$ of the directrix.

My question is are these well-known formulas for calculating the $a$, $b$ and $c$ coefficients given only information about the focus and directrix?

And what is the mathemathical proof for those formulas?

Update: how to calculate b:

Vertex form of parabola $x_{0}=-\frac{b}{2a}$

is simplified to $b = -2ax_{0}$,

and since $x_{0}$ of the vertex equals $x_{f}$ of the focus, $b = -2ax_{f}$,

and if $a = \frac{1}{2( y_{f} - y_{d})}$,

then $b = -2x_{f}.\frac{1}{2( y_{f} - y_{d})}=-\frac{2x_{f}}{2( y_{f} - y_{d})}$.

Update 3: a better explanation at calculation of $a$

According to Everything You (N)ever Wanted to Know About Parabolas there is a direct relation between $a$ and the distance between the focus and the vertex:

Focus and Directrix; Finally, it's important to note that the distance (d) from the vertex of the parabola to its focus is given by: $d = \frac{1}{4a}$

The distance between the focus and the directrx is two times this distance, so $d_{fd} = 2\frac{1}{4a} = \frac{1}{2a}$.

If we substitute the distance between the focus and directrix we get: $y_{f} - y_{d} = \frac{1}{2a}$.

$2a = \frac{1}{y_{f} - y_{d}}$

and

$a = \frac{1}{2(y_{f} - y_{d})}$

Which shows how $a$ is calculated.

Update 4: how to calculate c:

First we expand the vertex form to standard form:

$y = a(x – h)^2 + k$

becomes

$y = ax^2 -2ahx + (ah^2 + k)$,

where the last part in brackets plays the role of the $c$ coefficient.

So we consider $c$ equal to the part in the brackets,

$c = ah^2 + k$

Since $h$ is the $x$ of the vertex, which is equal to $x$ of the focus, we replace $h$ with $x_{f}$.

$c = ax_{f}^2 + k$

And since $k$ (the $y$ of the vertex) is at distance $\frac{1}{4a}$ from the $y$ of the focus, we replace $k$ with $(y_{f} - \frac{1}{4a})$:

$c = ax_{f}^2 + y_{f} - \frac{1}{4a}$

Then we replace $a$ with the value that was already computed for it $\frac{1}{2( y_{f} - y_{d})}$:

$c = \frac{1}{2( y_{f} - y_{d})}x_{f}^2 + y_{f} - \frac{1}{4\frac{1}{2( y_{f} - y_{d})}}$

which simplifies to:

$c = \frac{x_{f}^2}{2( y_{f} - y_{d})} + y_{f} - \frac{2( y_{f} - y_{d})}{4}$

which is almost the same as the formula from the code, except that $y_{f}$ is used instead of $y_{d}$. As $y_{d}$ is at the same distance from the vertex, as is $y_{f}$, the only difference is the sign.

• For $c$, isn’t it simply a matter of plugging in the already-computed values of $a$ and $b$ and setting $x=0$? – amd Mar 20 '18 at 17:11
• @amd It seems so, but I had to plug only $a$. – quasoft Mar 20 '18 at 18:41

Hint:

start from the definition:

a parabola is the locus of points $(x,y)$ that have the same distance from the directrix and from the focus.

Write this condition in your case and compare the result with your code.

If $P=(p_x,p_y)$ is the focus and $y=d_y$ is the equation of the directrix, the equation of the parabola is : $$(x-p_x)^2+(y-p_y)^2=(y-d_y)^2$$

• My math is too rusty, I don't remember what locus is. How the author of the code come up with these equations from that definition? – quasoft Mar 20 '18 at 13:25
• @quasoft: are you familiar with the formulae for the distance between two points, and the distance of a point from a line? You'll need to use both to assemble the parabola from its definition. – J. M. isn't a mathematician Mar 20 '18 at 14:51
• @quasoft: I added to my answer... – Emilio Novati Mar 20 '18 at 15:00
• Thanks, it was helpful – quasoft Mar 20 '18 at 15:46
• You are welcome! Don't forget to accept if you like the answer :) – Emilio Novati Mar 20 '18 at 16:26

Just look to the definition of a conic, how the present parabola special case is formed as a ratio ( =1 here) of distances

$$( x-x_f)^2+(y-y_f)^2 = (y-y_d)^2$$

The parabola vertex lies exact vertical mid-point of directix and parabola focus.