Given a focus and the directrix of a parabola, how do i get the general form of the parabola?
Example:
focus: $(12,10)$
Directrix: $3x + 2y = 22$
How do I get the formula:
$$A x^{2} + B xy + C y^{2} + D x + E y + F = 0$$
For the respective parabola?
I have already tried to look at the derivation of the above formula, but i can't figure it out, the reason is that the derivation is too general and complex, the derivation does not start from focus and directrix but from a plane and a cone.
On
Let focus be $F(12,10)$, a general point on parabola be $P(h,k)$ and $D$ be foot of perpendicular from $P$ to directrix.
For a parabola, $$\begin{align} PD^2&=PF^2\\ \frac{(3h+2k-22)^2}{13}&=(12-h)^2+(10-k)^2\\ 9h^2+4k^2+484+2(6hk-44k-66h)&=13\left(h^2+k^2-24h-20k+244\right)\\ 4h^2-12hk+9k^2-180h-172k+2688&=0\end{align}$$ Thus the equation of the parabola is $$\color{red}{4x^2-12xy+9y^2-180x-172y+2688=0}$$ NB: This satisfies the condition for coefficients for a parabola, i.e. $B^2(=12^2)=4AC(=4\cdot 4\cdot 9)$
Let the focus be at $(u,v)$, and the directrix have equation $ax+by+c=0$, where $a^2+b^2=1$.
We express equality of the squared distances:
$$(x-u)^2+(y-v)^2=(ax+by+c)^2.$$
Then, developing and grouping,
$$(1-a^2)x^2-2abxy+(1-b^2)y^2-2(ac+u)x-2(bc+v)y+u^2+v^2-c^2=0.$$
This is indeed a parabola because
$$(1-a^2)(1-b^2)-(ab)^2=1-a^2-b^2=0.$$
Another approach is to solve for a particular case such as the focus at the origin and the directrix parallel to the $x$ axis at distance $d$:
$$x^2+y^2=(y-d)^2$$ or
$$x^2+2dy-d^2=0$$and to transform the coordinates by arbitrary rotation and translation.
We get (using $c,s$ for the cosine and sine of the rotation angle)
$$(cx-sy+u)^2+2d(sx+cy+v)-d^2=0,$$
$$c^2x^2-2csxy+s^2y^2+2(cu+ds)x+2(cd-su)y+2dv+u^2-d^2,$$ which can be multiplied by an arbitrary constant.
Here again,
$$c^2\cdot s^2-(cs)^2=0.$$