# 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: T0$\left( -5,\frac{25}{4} \right)$, T1$\left( 1,\frac{1}{4} \right)$, T2$\left( 2,\frac{4}{4} \right)$, T3$\left( 3,\frac{9}{4} \right)$, T4$\left( 4,\frac{16}{4} \right)$, and so on. The tangent at T0 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 T1, T2, T3, T4, ... all cut the tangent at T0 into equal segments. The intersection points are shown as P1, P2, P3, P4, ....

In Properties of Parabolas, we learned that the intersection of two tangents at (a, ka²) and (b, kb²) is given by the formula $\left( \frac{a+b}{2},kab \right)$. Therefore, The intersection points of T0 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 D1,2, D2,3, D3,4, ... 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, kp²) and the other tangents be at (a, ka²), (a + d, k(a + d)²), (a + 2d, k(a + 2d)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, kp²) 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 I1 and I2 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, kp²) 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.