# Properties of Parabolas Revisited

# C o n c e p t s

This topic requires familiarity with the following concepts:

When I originally wrote about the properties of a parabola on the page Properties of Parabolas, I knew that there was more involved. I only looked at the intersection of two tangents. However, when you draw multiple tangents equally spaced apart, multiple intersections and multiple regions (quadrilaterals) are created. The discovery of the tangent-generated curve and that it is actually a parabola sparked me to revisit the parabola tangent intersections and the areas of the quadrilaterals. We'll look at two more properties here. The first is what I've called **equal dissection property of parabola tangents** for lack of better words.

**Equal Dissection Property of Parabola Tangents.** The tangents of a parabola mutually dissect each other into segments of equal length if the tangents are drawn at horizontally equally spaced points on the parabola.

## Equal Dissection Property of Parabola Tangents

Although awkwardly worded, all this means is that if you draw tangents at equally spaced points on the parabola (points equally spaced on the *x*-axis), then each tangent is cut into segments of equal length.

### Example

Before we prove this with variables, an example with real numbers will help. We’ll use the parabola $y=\frac{1}{4}{{x}^{2}}$. Draw the tangents at the following points: T_{0}$\left( -5,\frac{25}{4} \right)$, T_{1}$\left( 1,\frac{1}{4} \right)$, T_{2}$\left( 2,\frac{4}{4} \right)$, T_{3}$\left( 3,\frac{9}{4} \right)$, T_{4}$\left( 4,\frac{16}{4} \right)$, and so on. The tangent at T_{0} will be our "reference" tangent that will be dissected into equal segment. However, any one of the tangents can be used.

In the image above, tangents at T_{1}, T_{2}, T_{3}, T_{4}, ... all cut the tangent at T_{0} into equal segments. The intersection points are shown as P_{1}, P_{2}, P_{3}, P_{4}, ....

In Properties of Parabolas, we learned that the intersection of two tangents at (*a*, *ka*²) and (*b*, *kb*²) is given by the formula $\left( \frac{a+b}{2},kab \right)$. Therefore, The intersection points of T_{0} with the other tangents is given by:

${{P}_{1}}=\left( \frac{-5+1}{2},\frac{1}{4}\cdot (-5)(1) \right)=\left( -2,-\frac{5}{4} \right)$

${{P}_{2}}=\left( \frac{-5+2}{2},\frac{1}{4}\cdot (-5)(2) \right)=\left( -\frac{3}{2},-\frac{5}{2} \right)$

${{P}_{3}}=\left( \frac{-5+3}{2},\frac{1}{4}\cdot (-5)(3) \right)=\left( -1,-\frac{15}{2} \right)$

${{P}_{n-1}}=\left( \frac{-5+n-1}{2},\frac{1}{4}\cdot (-5)(n-1) \right)=\left( \frac{-5+n-1}{2},-\frac{5(n-1)}{4} \right)$

${{P}_{n}}=\left( \frac{-5+n}{2},\frac{1}{4}\cdot (-5)(n) \right)=\left( \frac{-5+n}{2},-\frac{5n}{4} \right)$

Let the distance between the points be D_{1,2}, D_{2,3}, D_{3,4}, ... Then, using the distance formula:

${{D}_{1,2}}=\sqrt{{{\left( \frac{-5}{4}+\frac{5}{2} \right)}^{2}}+{{\left( -2+\frac{3}{2} \right)}^{2}}}=\frac{\sqrt{29}}{4}$

${{D}_{2,3}}=\sqrt{{{\left( \frac{-5}{2}+\frac{15}{4} \right)}^{2}}+{{\left( -\frac{3}{2}+1 \right)}^{2}}}=\sqrt{\frac{25}{16}+\frac{4}{16}}=\frac{\sqrt{29}}{4}$

${{D}_{n-1,n}}=\sqrt{{{\left( \frac{-5+n}{2}-\frac{-5+n-1}{2} \right)}^{2}}+{{\left( -\frac{5n}{4}-\frac{-5(n-1)}{4} \right)}^{2}}}=\sqrt{{{\left( \frac{1}{2} \right)}^{2}}+{{\left( -\frac{5}{4} \right)}^{2}}}=\frac{\sqrt{29}}{4}$

Hence, for any two consecutive points, the distance is always the same.

## The General Length Formula

In the example, we found the distance for tangent points that are spaced 1 unit apart on the parabola $y=\frac{1}{4}{{x}^{2}}$. Let’s find a general formula for points spaced *d* units apart for the parabola $y={k}{x}^{2}}$. This will allow us to prove this theorem.

Let the "reference" tangent that is being dissected be at the point (*p*, *kp*²) and the other tangents be at (*a*, *ka*²), (*a* + *d*, *k*(*a* + *d*)²), (*a* + 2*d*, *k*(*a* + 2*d*)²), ..., [*a* + (*n* – 1)*d*, *k*(*a* + (*n* – 1)*d*)²], [*a* + *nd*, *k*(*a* + *nd*)²]. We can choose any of these points but to keep it general we’ll use the points at *a* + (*n* – 1)*d* and *a* + *nd* and (*p*, *kp*²) as the "reference" tangent.

The two intersection points of interest of these 3 tangents are:

${{I}_{1}}=\left( \frac{p+a+(n-1)d}{2},pa+kp(n-1)d \right)=\left( \frac{p+a+nd-d}{2},pa+kpnd-kpd \right)$

${{I}_{2}}=\left( \frac{p+a+nd}{2},kpa+kpnd \right)$

Let the length of these segments be *L*. The length between I_{1} and I_{2} can be found using the distance formula:

${{D}_{1,2}}=\sqrt{{{\left[ \left( \frac{p+a+nd-d}{2} \right)-\left( \frac{p+a+nd}{2} \right) \right]}^{2}}+{{\left[ \left( pa+kpnd-kpd \right)-\left( pa+kpnd \right) \right]}^{2}}}$

${{D}_{1,2}}=\sqrt{{{\left( -\frac{d}{2} \right)}^{2}}+{{\left[ -kpd \right]}^{2}}}=d\sqrt{\frac{1}{4}+{{k}^{2}}{{p}^{2}}}$

For a parabola $y = kx^2$, if tangents are drawn equally spaced apart horizontally on a parabola and the distance between them is *d*, then they divide a tangent drawn from (*p*, *kp*²) into equal lengths, where the length is given by the formula $L=d\sqrt{\frac{1}{4}+{{k}^{2}}{{p}^{2}}}$.

That is quite a formula for the length of the segments. Let’s verify that it works for our example above. The parabola was $y=\frac{1}{4}{{x}^{2}}$ and our segments were 1 unit apart so *d* = 1. The segment that was being dissected was at –5 so *p* = –5 and $k=\frac{1}{4}$. The length given by the formula is $L=1\sqrt{\frac{1}{4}+{{\left( \frac{1}{4} \right)}^{2}}{{(-5)}^{2}}}=\sqrt{\frac{1}{4}+\frac{25}{16}}=\sqrt{\frac{29}{16}}=\frac{\sqrt{29}}{4}$.

The formula verifies. We now turn to the area of the quadrilaterals that are formed by the tangents.