I was wondering if it is possible to derive a general form of a parabola given any focus and directrix.
So far all the materials I have come across only show the derivation for a parabola equation where the directrix is $x=c$ or $y=c$ for some constant $c$. And the only material I know that provides a general formula for a parabola is this article in wikipedia. But this relies on the general form of the conic equation.
I would like to derive the general equation of the parabola based on the definition of the parabola:
Let:
$d_1$ be the distance of a point on the parabola and its focus, $P(x_1,y_1)$
$d_2$ be the distance of a point on the parobola to its directrix, $y=mx+c$
$P(x,y)$ be any point on the parabola
So by definition of a parabola, $$\begin{align} d_1 &= d_2 \\ \sqrt{(x-x_1)^2 - (x-y_1)^2 } &= ??\end{align}$$
I can't proceed further as I don't know what to put for $d_2$ as all the textbook I consulted only have the directrix in the form of $x=c$ or $y=c$, which leads me to think that a derivation of the general parabola equation using this approach is impossible.
Please advise and provide the full steps if applicable.
Let:
So by definition of a parabola, $$d_1=\sqrt{(x−x_1)^2−(x−y_1)^2}=d_2$$.
$$(Y-y_1)=A(X-x_1)^2$$ where $A=$the degree and direction of parabola i.e. $-x^2$ is downward
$(y_1,x_1)$ is focus and directrix is $y=c=1/4A$
Derived from all points equidistant from focus to any $x,y$ and directrix, as formal definition of a parabola, from Pythagorean theorem
$$(X-a)^2+(Y-b)^2=(Y-c)^2$$
Separate $y$ values to one side and expand
$$(X-a)^2=Y^2-Y^2+2Yb-2Yc-b^2+c^2 =2Y(b-c)-(b^2-c^2)\\ (X-a)^2=2Y(b-c)-((b-c)(b+c))\\ (X-a)^2/(2(b-c))=Y-(b+c).$$ So $$A=\frac 12(b-c) \text{ and } x_1=a, y_1=b \text{ and }c=\text{ directrix}$$
Kind of one way to do I guess. Sure there is less convoluted solution to it but once you understand this you will get better.