Focal chords and the reflection property

What you will be able to do

• find the ends and the length of a focal chord;
• use the harmonic mean property of its two parts;
• find where the tangents at the ends of a chord meet;
• explain and use the reflection property.

A torch and a dish

Put a small bulb at the focus of a parabolic mirror. The beam leaves the torch in a straight line parallel to the axis. A satellite dish does the opposite: parallel signals arrive and gather at the receiver at the focus. Both use one property of the parabola. We prove it at the end of this lesson. First we study the chords that pass through the focus.

Focal chords

A chord that passes through the focus $F(a, 0)$ is a focal chord. Let one end be $P(at_1^2, 2at_1)$ and the other $Q(at_2^2, 2at_2)$. The slope of $PF$ is $\dfrac{2at_1}{a(t_1^2 - 1)} = \dfrac{2t_1}{t_1^2 - 1}$ and the slope of $QF$ is $\dfrac{2t_2}{t_2^2 - 1}$. They are equal exactly when

So if one end is at the parameter $t$, the other is at $-\dfrac{1}{t}$. The ends are $P(at^2, 2at)$ and $Q\left(\dfrac{a}{t^2}, -\dfrac{2a}{t}\right)$.

For $y^2 = 4x$ and the end $(4, 4)$ we have $t = 2$, so the other end is at $t = -\dfrac{1}{2}$, which is $\left(\dfrac{1}{4}, -1\right)$.

The length and the harmonic mean

The focal distances of the two ends are $SP = a(1 + t^2)$ and $SQ = a\left(1 + \dfrac{1}{t^2}\right)$. Their sum is the length of the chord:

The shortest focal chord is at $t = 1$, where $PQ = 4a$. This is the latus rectum. Every other focal chord is longer.

Now look at the reciprocals:

So $2a$, which is half the latus rectum, is the harmonic mean of $SP$ and $SQ$. Example. If $SP = 4$ and $SQ = 12$, then $\dfrac{1}{4} + \dfrac{1}{12} = \dfrac{1}{3}$, so $a = 3$ and the latus rectum is $12$.

The tangents at the ends of a focal chord

The tangents at $t_1$ and $t_2$ meet at the point $\left(at_1t_2,\ a(t_1 + t_2)\right)$. For a focal chord $t_1t_2 = -1$, so the meeting point is $\left(-a,\ a(t_1 + t_2)\right)$, which is on the directrix. The slopes of the tangents are $\dfrac{1}{t_1}$ and $\dfrac{1}{t_2}$, and their product is $\dfrac{1}{t_1t_2} = -1$. So the tangents at the ends of a focal chord are perpendicular and meet on the directrix.

Another fact of the same kind: the circle on a focal chord as diameter touches the directrix. The centre of the circle is the midpoint $M$ of the chord. Its distance from the directrix is the average of the two distances of the ends, which is $\dfrac{SP + SQ}{2}$, the radius.

The reflection property

Let $T$ be the point where the tangent at $P(at^2, 2at)$ meets the axis. From the last lesson $T = (-at^2, 0)$. The focus is $F(a, 0)$. Then $FT = a + at^2 = a(1 + t^2)$, and the focal distance is $FP = a(1 + t^2)$. So

and triangle $FPT$ is isosceles. Its base angles are equal: the angle between the tangent and $PF$ equals the angle at $T$ between the tangent and the axis. The line $PR$ drawn parallel to the axis makes the same angle with the tangent as the axis does, as alternate angles. So the tangent makes equal angles with the line $PF$ and with the line parallel to the axis.

A ray that obeys the law of reflection makes equal angles with the tangent before and after. So a ray travelling parallel to the axis reflects through the focus, and a ray from the focus reflects parallel to the axis. Physics books call this the focusing property of a parabolic mirror. Telescopes, car headlamps, dish antennas and solar cookers all use it. A spherical mirror does not focus exactly, which is why large mirrors are paraboloids.

A worked example

On $y^2 = 12x$ take the focal chord with $t = 2$, so $a = 3$.

1. The ends. $P = (3 \times 4,\ 2 \times 3 \times 2) = (12, 12)$. The other end is at $t = -\dfrac{1}{2}$: $Q = \left(3 \times \dfrac{1}{4},\ 2 \times 3 \times \left(-\dfrac{1}{2}\right)\right) = \left(\dfrac{3}{4}, -3\right)$.
2. The length. $PQ = 3\left(2 + \dfrac{1}{2}\right)^2 = 18.75$.
3. The two parts. $SP = 3(1 + 4) = 15$ and $SQ = 3\left(1 + \dfrac{1}{4}\right) = 3.75$. They add up to $18.75$.
4. The harmonic check. $\dfrac{1}{15} + \dfrac{1}{3.75} = \dfrac{1}{15} + \dfrac{4}{15} = \dfrac{1}{3}$, which is $\dfrac{1}{a}$.
5. The tangents. At $t = 2$ the slope is $\dfrac{1}{2}$, and at $t = -\dfrac{1}{2}$ the slope is $-2$. The product is $-1$. They meet on $x = -3$ at $\left(-3,\ 3\left(2 - \tfrac{1}{2}\right)\right) = (-3, 4.5)$.

Practise with new numbers

Try it in Python

The program takes the parameter $t$ of one end of a focal chord and finds the other end, the length, the two parts and the point where the two tangents meet.

Common mistakes

• Using $t_1t_2 = 1$. For a focal chord the product is $-1$.
• Forgetting that the chord length is a square. It is $a\left(t + \dfrac{1}{t}\right)^2$, not $a\left(t^2 + \dfrac{1}{t^2}\right)$.
• Forgetting that the tangents meet on the directrix only for a focal chord. For other chords they meet elsewhere.
• Reading the reflection property as an angle with the axis. The equal angles are measured against the tangent.

Remember these

• Ends of a focal chord: $t$ and $-\dfrac{1}{t}$.
• Length: $a\left(t + \dfrac{1}{t}\right)^2 \ge 4a$, with equality for the latus rectum.
• $\dfrac{1}{SP} + \dfrac{1}{SQ} = \dfrac{1}{a}$.
• The tangents at the ends of a focal chord are perpendicular and meet on the directrix.
• A ray parallel to the axis reflects through the focus, because $FT = FP$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A focal chord of $y^2 = 4x$ has one end at $(4, 4)$. Find the other end.

2. The two parts of a focal chord of a parabola are $4$ and $12$. Find the length of the latus rectum.

3. What is the shortest focal chord of $y^2 = 12x$?

4. Why are the tangents at the ends of a focal chord perpendicular?

5. A ray travels parallel to the axis of a parabolic mirror and hits the mirror. Where does it go after the reflection?

References

• NCERT, Mathematics for Class XI, Chapter 10 (Conic Sections), the section on the parabola.
• OpenStax, Precalculus 2e, Chapter 12 (Analytic Geometry), the section on the parabola, which explains the reflection property of a parabolic mirror.