The parabola: equal distances to a focus and a directrix

Parabola

Why is the focal radius of y² = 2px just x + p/2, and how long is a chord through the focus?

Drag a point along y² = 2px and compare its distance to the focus with its distance to the directrix: they are always equal. Then turn a chord through the focus and see why its length is x₁ + x₂ + p = 2p / sin²α, shortest when it stands upright.

Drag the ringed points on the board.

High school
Focus and directrix
Standard equation y² = 2px
Focal radius
Focal chord and latus rectum

p

2.0

y_P

2.0

\alpha\,(^\circ)

latus rectum

|PF| = |PD| = x + \tfrac{p}{2}

$|PF|$: 2.000
$|PD| = x_P + \tfrac{p}{2}$: 2.000
$|AF|$: 4.000
$|BF|$: 1.333
$|AB|$: 5.333
$\dfrac{2p}{\sin^2\alpha}$: 5.333
$\dfrac{1}{|AF|} + \dfrac{1}{|BF|}$: 1.000

Guided tour

1Same distance to F and to the line
With $p=2$ the focus is $F=(1,0)$ and the directrix is $x=-1$ . At $P=(1,2)$ : $|PF|=2$ and the distance to the directrix is $|PD|=2$ . That equality is the whole definition of the parabola.
2A distance you can read off x
Drag $P$ up to $y_P=4$ , so $x_P=4$ . Computing $|PF|=\sqrt{(4-1)^2+4^2}=5$ takes a square root, but $|PD|=x_P+\frac{p}{2}=4+1=5$ does not. Swapping one distance for the other is the main trick with parabolas.
3A chord through the focus
The amber chord meets the parabola at $A$ and $B$ at $\alpha=60^\circ$ . Each focal radius equals its own drop to the directrix, so $|AB|=x_A+x_B+p$ . Here $|AF|=4$ , $|BF|=1.333$ and $|AB|=5.333$ , matching $\frac{2p}{\sin^2\alpha}=\frac{4}{3/4}$ .
4Upright: the shortest focal chord
At $\alpha=90^\circ$ the chord is the latus rectum: $|AF|=|BF|=p=2$ and $|AB|=2p=4$ , the shortest of all, since $\sin^2\alpha\le1$ . Turn it to $30^\circ$ and it grows to $16$ . Through all of this, $\frac{1}{|AF|}+\frac{1}{|BF|}$ stays at $\frac{2}{p}=1$ .

Explanation

1
Definition and equation
A parabola is the set of points equally far from a fixed point $F$ (the focus) and a fixed line $l$ (the directrix) not through $F$ . Let $p$ be the distance from $F$ to $l$ , and put $F=\left(\frac p2,0\right)$ , $l:\ x=-\frac p2$ . Then
$\sqrt{\left(x-\tfrac p2\right)^2+y^2}=x+\tfrac p2$
Squaring both sides, the $x^2$ and $\frac{p^2}{4}$ terms cancel and leave
$y^2=2px\qquad(p>0)$
2
Focal radius
For $P(x_0,y_0)$ on $y^2=2px$ :
$|PF|=x_0+\frac p2$
This is not a new fact, only the definition read the other way: the distance to the focus is replaced by the distance to the directrix, which is a horizontal length.
3
Length of a focal chord
Add the two focal radii: $|AB|=|AF|+|BF|=x_1+x_2+p$ .
For the angle, write the chord as $x=\frac p2+s\cos\alpha$ , $y=s\sin\alpha$ and substitute: $s^2\sin^2\alpha-2ps\cos\alpha-p^2=0$ , whose roots are $s=\dfrac{p}{1-\cos\alpha}$ and $s=-\dfrac{p}{1+\cos\alpha}$ . So
$|AB|=\frac{p}{1-\cos\alpha}+\frac{p}{1+\cos\alpha}=\frac{2p}{\sin^2\alpha}$
and $\dfrac{1}{|AF|}+\dfrac{1}{|BF|}=\dfrac{(1-\cos\alpha)+(1+\cos\alpha)}{p}=\dfrac2p$ for every direction.
4
Why dishes are parabolic
Since $|PF|=|PD|$ , triangle $PFD$ is isosceles, and the line through $P$ that bisects $\angle FPD$ is the perpendicular bisector of $FD$ . Every other point $Q$ of that line is equally far from $F$ and $D$ too, but $Q$ is farther from $D$ than from the directrix, so $Q$ lies outside the parabola: the line touches the curve only at $P$ , which makes it the tangent. So the tangent bisects the angle between $PF$ and the horizontal through $P$ . A ray coming in parallel to the axis therefore reflects through the focus, and light from a bulb at the focus leaves as a parallel beam. Satellite dishes, car headlights and solar cookers all use this.

Check yourself

Q01

A point $P$ on $y^2=8x$ has $x=3$ . What is $|PF|$ ?

Q02

A chord of $y^2=4x$ through the focus has $x_1+x_2=6$ . How long is it?

Q03

What is the shortest chord of $y^2=2px$ through the focus?

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