Understanding general case of parabola equation

Question

I'm working on understanding how to find the equation of a parabola using directrix and focus. I understand the formulas that use the pythagorean theorem/distance formula to equate the distance from a point of the parabola $P = (x, y)$ and the directrix to the distance of said point to the focus. I also got how it works for parabolas opening left/right or opening up/down.

Now I was wondering, how does it work in the general case, for example, if the parabola is opening to the "top-right" or such.

On wikipedia I found the general formula for this:

If the focus is $F=(f_{1},f_{2})$ and the directrix $ax+by+c=0$ one gets the equation

$$\frac{(ax+by+c)^2}{a^2+b^2} = (x-f_1)^2+(y-f_2)^2$$

(The left side of the equation uses the Hesse normal form of a line to calculate the distance $|Pl|$.)

With the right side I'm familiar, it calculcates delta $x$ and delta $y$ and we equal that to the left side, the distance of some point $(x, y)$ and the directrix.

But I found the left side is giving me trouble, I read the linked wiki article on the Hesse normal form, but it didn't make it very clear to me how the distance there is calculated.

Can anyone explain more how the distance on the left hand side of the equation is calculated? What is $a$ and $b$ in this case?

Answer 1

Let $P(x_1,x_2)$ and $\ell : ax+by+c=0$.

If we want to find the distance between $P$ and the line $\ell$, then let $H(\alpha,\beta)$ be a point on $\ell$ such that $PH\perp\ell$.

We can write $$PH^2=(\alpha-x_1)^2+(\beta-y_1)^2\tag1$$ Since $H$ is on $\ell$, $$a\alpha+b\beta+c=0\tag2$$ Since $PH\perp\ell$, $$\frac{\beta-y_1}{\alpha-x_1}=\frac ba$$ implying $$\frac{\alpha-x_1}{a}=\frac{\beta-y_1}{b}$$ If we set $$\frac{\alpha-x_1}{a}=\frac{\beta-y_1}{b}=k$$ we have $$\alpha=ak+x_1,\qquad \beta=bk+y_1\tag3$$ Substituting $(3)$ into $(2)$ and solving it for $k$ gives $$k=\frac{ax_1+by_1+c}{a^2+b^2}\tag4$$ It follows from $(1)(3)(4)$ that $$PH^2=(a^2+b^2)k^2=\frac{(ax_1+by_1+c)^2}{a^2+b^2}$$ It follows that the distance between $P$ and the line $\ell : ax+by+c=0$ is given by $$PH=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$$

Answer 2

Suppose we want to find the parpendicular distance of $L:~ax+by+c=0$ from $P:~(x_0,y_0)$. Note that the slope of $L$ is $-\frac ab$. Therefore slope of a line parpendicular to $L$ will be $\frac ba$. There equation if the line parpendicular to $L$ passing through $P$ will be $$L':~y-y_0=\frac ba(x-x_0)$$ Now find the point $P'$ where the line $L'$ intersects with $L$, by solving the corresponding linear equations for $L$ and $L'$. Then the distance between the points $P$ and $P'$ will give you the required distance. Carry out these calculations to get the given formula.

Answer 3

The vertical distance of a point to the axis $x$ is obviously $|y|$.

Now if you rotate the plane by an angle $\theta$, the ordinate is transformed by

$$y'=x\cos\theta+y\sin\theta.$$

And if you translate the plane to another origin,

$$y''=y'+c=x\cos\theta+y\sin\theta+c.$$

All the points at distance $0$ from the line thus fulfill the condition

$$x\cos\theta+y\sin\theta+c=0,$$ (this is the implicit equation of the line) and those at distance $d$,

$$\left|x\cos\theta+y\sin\theta+c\right|=d.$$

Instead of the angle $\theta$, you can consider two coefficients $a,b$ and write

$$\cos\theta=\lambda a,\\\sin\theta=\lambda b.$$ By elimination of $\lambda$,

$$\cos\theta=\frac a{\sqrt{a^2+b^2}},\\\sin\theta=\frac b{\sqrt{a^2+b^2}} b.$$

Then

$$d^2=\left(\frac{ax+by}{\sqrt{a^2+b^2}}+c\right)^2$$ or

$$d^2=\frac{(ax+by+c')^2}{a^2+b^2}.$$