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:


$d_1$ 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.

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    $\begingroup$ Do you know linear algebra? If so, the easiest way would be to find the equation of the parabola when the directrix is of the form $x=c$ (or $y=c$) and rotate the coordinate system. $\endgroup$ – Étienne Bézout Oct 10 '13 at 14:33
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    $\begingroup$ In analytic geometry one studies the following formula for the point-line distance $$d=\frac {|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$ $\endgroup$ – Tony Piccolo Oct 10 '13 at 14:42
  • $\begingroup$ @ÉtienneBézout I do know linear algebra. But like to solve it using this approach first. So I gather from your comment that this approach is feasible but tedious. Which part of it is tedious? $\endgroup$ – mauna Oct 10 '13 at 14:44
  • $\begingroup$ @mauna I recall attempting your approach a few years ago in my linear algebra course, and I think it resulted in some equations which were rather tedious to solve. Also, if you try to take a general directrix on the form $y=mx+c$ you will not cover the case of vertical directrices. In your approach, it is probably best to write the directrix on the form $ax+by+c=0$ and follow Tony Piccolo's suggestion. $\endgroup$ – Étienne Bézout Oct 10 '13 at 14:50
  • $\begingroup$ I would recommend equating the squares of the distances, which gets rid of the square roots and absolute values. Also the equation with the square root and two question marks looks a bit mixed up to me - I'm not sure what the left-hand side is supposed to be. $\endgroup$ – Mark Bennet Oct 10 '13 at 15:59

$$d_1=\sqrt{(x-x_1)^2+(y-y_1)^2}$$ And $$d_2=\frac{|y-mx+c|}{\sqrt{1+m^2}}$$ You can form the equation of Parabola now, but as you were unsure about second, I'll help you prove it:

As we are measuring perpendicular distance, take the line perpendicular to $y=mx+c$ passing through $(x_0,y_0)$ and the foot of perpendicular on line $(\alpha,\beta)$,i.e.$$(\beta-y_0)=\frac{-1}m(\alpha-x_0)$$ Or, $$m(\beta-y_0)+(\alpha-x_0)=0$$ Squaring, $$m^2(\beta-y_0)^2+(\alpha-x_0)^2=-2m(\alpha-x_0)(\beta-y_0)\tag1$$ Now consider, $$(m(\alpha-x_0)-(\beta-y_0))^2=m^2(\alpha-x_0)^2+(\beta-y_0)^2-2m(\alpha-x_0)(\beta-y_0)$$ Or $$m^2(\alpha-x_0)^2+(\beta-y_0)^2-(m(\alpha-x_0)-(\beta-y_0))^2=2m(\alpha-x_0)(\beta-y_0)\tag2$$ Adding (1) and (2), $$m^2(\beta-y_0)^2+(\alpha-x_0)^2+m^2(\alpha-x_0)^2+(\beta-y_0)^2=(m(\alpha-x_0)-(\beta-y_0))^2$$ Or [Use $c=\beta-m\alpha$ and rearrange] $$(m^2+1)((\beta-y_0)^2+(\alpha-x_0)^2)=(y_0-mx_0-c)^2$$ So distance from line is: $$d=\sqrt{(\beta-y_0)^2+(\alpha-x_0)^2}=\frac{|(y_0-mx_0-c)|}{\sqrt{m^2+1}}$$ Note: For a line $ax+by=c$, put $m=-\frac ab$ to get: $$d=\frac {|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$

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$d_1$ be the distance of a point on the parabola and its focus, $P(x1,y1)$ $d_2$ 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, $$d_1=\sqrt{(x−x_1)^2−(x−y_1)^2}=d_2$$.

$$(Y-y_1)=A(X-x_1)^2$$ 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


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.

enter image description here

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The equation of the parabola using your notation is

\begin{multline} x^2 + m^2\,y^2 + 2m\,xy \\ {}- 2\bigl(x_1(1+m^2)+mc\bigr)\,x {}- 2\bigl(y_1(1+m^2)-c\bigr)\,y \\ + \bigl((x_1^2+y_1^2)(1+m^2)-c^2\bigr) =0 \end{multline}

In the derivation below I'll change the notation to avoid the indices and to make the situation more generic with respect to the equation of the line. I'll write the focus as $F=(x,y)$ and the directrix using the equation $aX+bY+c=0$ (as Étienne Bézout suggested). So $(X,Y)$ denotes a generic point against which a certain equation can be tested. My $x$ is your $x_1$, my $y$ your $y_1$, and since your line is $mx-y+c=0$, my $a$ is your $m$, my $b$ is $-1$ and my $c$ agrees with yours.

The vector $(a,b)$ denotes the direction orthogonal to the directrix. So the point on the directrix closest to a given point $(X,Y)$ is $(X,Y)+\lambda(a,b)=(X+\lambda a,Y+\lambda b)$ with $\lambda$ chosen such that it satisfies the equation of the line, namely

\begin{align*} a(X+\lambda a)+b(Y+\lambda b)+c&=0 \\ \lambda&=-\frac{a\,X+b\,Y+c}{a^2+b^2} \end{align*}

If you had the equation of the line scaled such that $a^2+b^2=1$, then this $\lambda$ would already be the length $d_2$. Otherwise, you have to multiply it by $\lVert(a,b)\rVert=\sqrt{a^2+b^2}$ (since your distance vector is $\lambda(a,b)$ so it has length $d_2=\lambda\lVert(a,b)\rVert$). You will find the length to be (like Tony Piccolo already stated):


Now that absolute value and that square root is ugly, so I'd square both sides (as suggested by Mark Bennet):

\begin{align*} d_1^2 &= d_2^2 \\ (X-x)^2+(Y-y)^2 &= \frac{(a\,X+b\,Y+c)^2}{a^2+b^2} \\ \bigl(a^2+b^2\bigr)\bigl((X-x)^2+(Y-y)^2\bigr) &= (a\,X+b\,Y+c)^2 \\ \bigl(a^2+b^2\bigr)\bigl((X-x)^2+(Y-y)^2\bigr) - (a\,X+b\,Y+c)^2 &= 0 \end{align*}

If you expand this and then collect terms with common powers of $X,Y$, you end up with the equation given initially, except for the change in notation:

\begin{multline} b^2\,X^2 + a^2\,Y^2 - 2ab\,XY \\ {}- 2\bigl(x(a^2+b^2)+ac\bigr)\,X {}- 2\bigl(y(a^2+b^2)+bc\bigr)\,Y \\ + \bigl((x^2+y^2)(a^2+b^2)-c^2\bigr) =0 \end{multline}

I first dealt with this whole situation in a different context, obtaining the formula using a different description of the parabola. But it's nice to see it confirmed like this.

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