# 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

The image on the right shows a parabola $$y=\frac{1}{2}{{x}^{2}}$$ with tangents drawn at the blue points that are evenly spaced 1 unit apart. We will take the red tangent as the reference tangent and the other tangents are dissecting this tangent. The reference tangent is at the point (–4, 8).

Therefore, the lengths P1P2, P2P3, P3P4, and so on are equal in length. The formula for the length is $$L=\frac{d}{2}\sqrt{1+(2kp)^2}$$, where d is the horizontal distance between the points of tangency, and p is the x value of the point of tangecy of the reference tangent. For our example, d = 1 and p = –4. (We will prove this formula below.) Substituting these value, we get $$L = \frac{1}{2}\sqrt{1+(2\cdot \frac{1}{2} \cdot 4)^2} = \frac{1}{2}\sqrt{17} \approx 2.062$$. This can be confirmed with the value calculated by Geogebra.

Before we prove this formula, let’s look at another example.

### Example: Manual Calculation

Consider the parabola $$y = \frac{1}{4}x^2$$ and a reference tangent, T0, at (–5, 25/4). We will dissect this tangent with tangents at x = 1, 2, 3, 4, .... n – 1, n. We’ll call these tangents T1, T2, T3, etc. Points P1, P2, P3, etc. will be the intersection points on T0 by the respective tangents. Image below will help visualize our set up.

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:

$$P_1 = \left( \frac{-5+1}{2},\frac{1}{4}\cdot (-5)(1) \right) = \left( -2,-\frac{5}{4} \right)$$

$$P_2 = \left( \frac{-5+2}{2},\frac{1}{4}\cdot (-5)(2) \right) = \left( -\frac{3}{2},-\frac{5}{2} \right)$$

$$P_3 = \left( \frac{-5+3}{2},\frac{1}{4}\cdot (-5)(3) \right) = \left( -1,-\frac{15}{4} \right)$$

$$P_{n-1} = \left( \frac{-5+n-1}{2},\frac{1}{4}\cdot (-5)(n-1) \right) =$$ $$\left( \frac{n-6}{2},-\frac{5(n-1)}{4} \right)$$

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

Let the distance between the points be D1,2, D2,3, D3,4, ... Then, using the distance formula:

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

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

$$D_{n-1,n} = \sqrt{\left( \frac{n-5}{2}-\frac{n-6}{2} \right)^2 + \left( -\frac{5n}{4}-\frac{-5(n-1)}{4} \right)^2} =$$ $$\sqrt{\left( \frac{1}{2} \right)^2 + \left( -\frac{5}{4} \right)^2} = \frac{\sqrt{29}}{4}$$

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

Now, let’s use the formula $$L=\frac{d}{2}\sqrt{1+(2kp)^2}$$ to find the length. Our parabola is $$y = \frac{1}{4}x^2$$. So, k = 1/4, d = 1 because the tangents are 1 unit apart, and p = –5 because the reference tangent is at x = –5. Therefore, the length is $$L=\frac{1}{2}\sqrt{1+[(2)(1/4)(-5)]^2} =$$ $$\frac{1}{2}\sqrt{1 + (-5/2)^2} = \frac{\sqrt{29}}{4}$$

Geogebra calculated the length to be 1.346, which is approximately equal to our exact value.

## 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 formula.

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 need to select 1 reference tangent and 2 other tangents that create 2 points of intersection with the reference tangent.

We can choose any three of these points, but to keep it general, we’ll use the points at x = a + (n – 1)d and a + nd, and we will use the tangent at (p, kp²) as the “reference” tangent.

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

$$I_1 = \left( \frac{p+a+(n-1)d}{2},kp[a+(n-1)d] \right)=$$ $$\left( \frac{p+a+nd-d}{2},kpa+kpnd-kpd \right)$$

$$I_2 = \left( \frac{p+a+nd}{2},kpa+kpnd \right)$$

We will find the length, L, between I1 and I2 using the distance formula:

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

$$D_{1,2} = \sqrt{\left( -\frac{d}{2} \right)^2 + (-kpd )^2} =$$ $$d\sqrt{\frac{1}{4}+k^2}{p^2} = \frac{d}{2}\sqrt{1+(2kp)^2}$$

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 $$L=\frac{d}{2}\sqrt{1+(2kp)^2}$$.

That is quite a formula for the length of the segments.

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

## Infinite Constant Area Regions

Suppose a parabola y = ax2 has 4 tangents at A, B, C, and D. Let the horizontal distance between A and B and the horizontal distance between C and D equal d. Let the horizontal distance between A and C equal D. Then, the area, A, of the quadrilateral created by the intersection points of the 4 tangents is contant and equal to $$A=\frac{1}{2}ad^2D$$.

In the figure above, the parabola is $$y=\frac{1}{2}x^2$$. Tangents at A, B, C, and D have been drawn. Horizontal distance between A to B and C to D is 1 and the horizontal distance between A to C is 3. So the area of the shaded quadrilateral EFGH is $$A = \frac{1}{2}\cdot\frac{1}{2}(1)^2(3) = \frac{3}{4}$$. As can be seen in the sketch created in Geogebra, the area calculated by Geogebra is also the same.

In the following graph, I have shifted all the points while keeping d and D the same. (Point D does not fall in the region shown.) Again, Geogebra calculated area is still 0.75 units.

### Proof

Let the 4 points be (p, ap2), (p + d, a(p + d)2), (q, aq2), (q + d, a(q + d)2). The intersections of these 4 lines are the following 4 coordinates:

(i) $$E = (x_{0}, y_{0}) = \left ( \frac{1}{2}(p+q+d), a(p+d)q\right )$$

(ii) $$H = (x_{1}, y_{1}) = \left ( \frac{1}{2}(p+q+2d), a(p+d)(q+d)\right )$$

(iii) $$G = (x_{2}, y_{2}) = \left ( \frac{1}{2}(p+q+d), ap(q+d)\right )$$

(iv) $$F = (x_{3}, y_{3}) = \left ( \frac{1}{2}(p+q), apq\right )$$

The area of a quadrilateral defined by 4 points is given by the formula: $$2A = (x_{2}- x_{0})(y_{3}- y_{1}) - (x_{3}- x_{1})(y_{2}- y_{0})$$. Making the appropriate substitutions, we have the following formula for the area of this quadrilateral:

(i) $$2A=\left [ \frac{1}{2}(p+q+d) - \frac{1}{2}(p+q+d)) \right ]\left [ apq - a(p+d)(q+d) \right ]-$$ $$\left [ \frac{1}{2}(p+q) - \frac{1}{2}(p+q+d)) \right ]\left [ ap(q+d) - aq(p+d) \right ]$$

(ii) $$2A=-(d)(apq+apd-apq-aqd) = ad^2(q-p)$$

If we let the distance between p and q (which is the horizontal distance between A to C) be D, then we get a nice formula for the area of the quadrilateral formed by 4 tangents to a parabola:

The area of a quadrilateral formed by the intersection of 4 tangents at A, B, C, and D is given by the formula: $$2A = ad^2D$$, where d is the horizontal distance between A to B and C to D, and D is the horizontal distance between A and C.

If many of these tangents were drawn equally spaced apart, then the areas of the quadrilaterals will all be equal.

The image above shows the contant area regions created by the parabola’s tangents all equally space apart on the horizontal axis. These areas were calculated by Geogebra. Even the triangles created by the tangents close to the parabola have equal areas, and the area between two tangents and the parabola also have constant areas. For tangents of a parabola $$y = \frac{1}{4}x^2$$ (pictured above) that are 1 units apart, the areas progressively increase by 0.125 units squared.