# C o n c e p t s

This topic requires familiarity with the following concepts:

The parabola has some interesting properties compared to higher-powered curves. Its derivative is linear so many of its properties reduce to constant values. We'll explore intersection point of two tangents, constant area regions, and reveal a concealed rhombus among points of interest.

## Intersection of Two Tangents

The first property I want to explore is the intersection of two tangents of the parabola. We will define the parabola as one with its vertex at the origin and has the equation y = kx². Parabolas whose vertex is not at the origin also possess the same properties. A study of Intercept Formulas & Newton's Method may assist you a bit here. The image on the right shows a parabola with two tangents intersecting at some point.

The y-intercept formula of a parabola defined by y = kx² is yyi = -kx². Its derivative (which gives the slope of the tangent) is y' = 2kx. So, a line tangent to the parabola at point p has the equation y = -2kpx - kp².

A tangent to a parabola y = kx² at point (p, kp²) is defined by the equation y = -2kpx - kp².

A line which is tangent at point a and b would have the equations y = -2kax - ka² and y = -2kax - ka², respectively. Set these equations equal to each other and solve for x to find their intersection point.

(i) $(2ka)x-k{{a}^{2}}=(2kb)x-k{{b}^{2}}$

(ii) $x(2ka-2kb)=k({{a}^{2}}-{{b}^{2}})$

(iii) $x=\frac{k(a+b)(a-b)}{2k(a-b)}$

(iv) $x=\frac{a+b}{2}$

Now find y by substitution into either of the line equations.

(vi) $y=(2ak)\left( \frac{a+b}{2} \right)-k{{a}^{2}}=kab$

The intersection point of two tangents of a parabola y = kx² at (a, ka²) and (b, kb²) is $\left( \frac{a+b}{2},kab \right)$. The x value of the intersection point is the arithmetic mean of the x values of the two tangents and the y value of the intersection point is the geometric mean of the y values of the two tangents.

It's easy to see that the x value is the arithmetic mean. The geometric mean is the square root of the product: $\sqrt{k{{a}^{2}}\cdot k{{b}^{2}}}=\sqrt{{{k}^{2}}{{a}^{2}}{{b}^{2}}}=kab$. Here, we have to take the absolute values as the actual intersection point can be negative.

## Constant Area Regions of a Parabola

The tangents and secants of parabolas create regions that have constant areas. The only variable is the horizontal distance between the two points and k, the coefficient of the parabola.

Region 1

Two tangents to a parabola create an enclosed region of constant area if the horizontal distance between the two tangents is constant. This area is shown in orange in the image. Consider line l tangent at point (a, ka²) and line m tangent at point (b, kb²) to a parabola defined by equation y = kx². The bounded area equals the area under the parabola from a to b "minus" the area of the two trapezoids. Depending on where the tangents are, the trapezoids could be triangles. Also, the trapezoids could fall below the x-axis. Hence, the area may be an addition instead of subtraction. We don't have to worry about these different scenarios as the math takes care of all this.

The area, Ap, from a to b under the curve is $A_{p}=\int\limits_{a}^{b}{x^2\text{ }dx}=\frac{k({{b}^{3}}-{{a}^{3}})}{3}$. The area of the first trapezoid, A1, is ${{A}_{1}}=\frac{1}{2}\left[ \frac{a+b}{2}-a \right](kab+k{{a}^{2}}\text{)}=\frac{(b-a)(kab+k{{a}^{2}})}{4}$. The area of the second trapezoid, A2, is ${{A}_{2}}=\frac{\text{(}b-a\text{)(}kab+k{{b}^{2}}\text{)}}{4}$. The sum of the areas of the trapezoids is ${{A}_{1}}+{{A}_{2}}=\frac{\text{(}b-a\text{)}[k\text{(}{{a}^{2}}+2ab+{{b}^{2}}\text{) }\!\!]\!\!\text{ }}{4}$. Now subtracting this sum from Ap gives us the area enclosed by the parabola and its two tangents, A:

(i) $A=\frac{4k\text{(}{{b}^{3}}-{{a}^{3}}\text{)}-3k\text{(}b-a\text{)(}{{a}^{2}}+2ab+{{b}^{2}}\text{)}}{12}=\frac{4k\text{(}b-a\text{)(}{{b}^{2}}+ab+{{a}^{2}}\text{)}-3k\text{(}b-a\text{)(}{{a}^{2}}+2ab+{{b}^{2}}\text{)}}{12}$

Factoring and simplifying:

(ii) $A=\frac{k\text{(}b-a\text{)(}4{{b}^{2}}+4ab+4{{a}^{2}}-3{{a}^{2}}-6ab-3{{b}^{2}}\text{)}}{12}=\frac{k\text{(}b-a\text{)(}{{b}^{2}}+{{a}^{2}}-2ab\text{)}}{12}=\frac{k{{\text{(}b-a\text{)}}^{3}}}{12}$

If we let d be the distance from a to b. Then a + d = b. Replace b with a + d.

(v) $A=\frac{k{{(b-a)}^{3}}}{12}=\frac{k{{(a+d-a)}^{3}}}{12}=\frac{k{{d}^{3}}}{12}$

The area, A bounded by a parabola and two tangent lines is $A=\frac{k{{d}^{3}}}{12}$, where d is the horizontal distance between the two points of tangency.

Region 2

Now we'll find another simple formula for the area bound by a secant line and the parabola. Let the secant line pass through the points (a, ka²) and (b, kb²). The area of the region will be that of the area under the secant line form a to b minus the area under the parabola from a to b. The equation of the secant line is:

(i) $y-k{{a}^{2}}=\frac{k{{b}^{2}}-k{{a}^{2}}}{b-a}(x-a)$

(ii) $y-k{{a}^{2}}=\frac{k(b+a)(b-a)}{(b-a)}(x-a)$

(iii) $y=k(a+b)x-ka(a+b)+k{{a}^{2}}$

(iv) $y=k(a+b)x-kab$

To find the area enclosed by the secant and the parabola, we can use integration in one step to find the area:

(i) $A=\int\limits_{a}^{b}{\text{ }k\text{(}a+b\text{)}x-kab-k{{x}^{2}}\text{ }dx}\text{ }=\left[ \frac{k\text{(}a+b\text{)}{{x}^{2}}}{2}-kabx-\frac{k{{x}^{3}}}{3} \right]_{a}^{b}$

(ii) $A=\frac{k}{6}[3(a+b)({{b}^{2}}-{{a}^{2}})-6ab(b-a)-2({{b}^{3}}-{{a}^{3}})] =\frac{k}{6}[3(a+b)(a+b)(b-a)-6ab(b-a)-2(b-a)({{b}^{2}}+ab+{{a}^{2}})]$

Now let d equal the distance from a to b. Then b = a + d.

(iii) $A=\frac{k}{6}[3(a+a+d)(a+a+d)(a+d-a)-6a(a+d)(a+d-a)-2(a+d-a)[{{(a+d)}^{2}}+a(a+d)+{{a}^{2}}]]$

(iv) $A=\frac{k}{6}[3d{{(2a+d)}^{2}}-6ad(a+d)-2d({{a}^{2}}+2ad+{{d}^{2}}+{{a}^{2}}+ad+{{a}^{2}}]$

(v) $A=\frac{k}{6}[12d{{a}^{2}}+12a{{d}^{2}}+3{{d}^{3}}-6d{{a}^{2}}-6a{{d}^{2}}-6d{{a}^{2}}-6a{{d}^{2}}-2{{d}^{3}}]$

(vi) $A=\frac{kd^{3}}{6}$

The area, A, bound by a secant and a parabola is dependent on the distance between the two point the secant passes on the parabola and the coefficient of the parabola and is given by $\frac{kd^3}{6}$!

Note that the area bound by a secant passing through (a, ka²) and (b, kb²) is twice the area bounded by a parabola and tangents at these points.

## The XY-Intercepts

As we saw earlier, the y-intercept of a tangent line to a parabola is given by yyi = -kx². This means that the y value of the point of tangency is the same as the y-intercept for a parabola with vertex on the origin.

The x-intercept is given by ${{y}_{xi}}=x-y\frac{dx}{dy}=x-(k{{x}^{2}})\left( \frac{1}{2kx} \right)=x-\frac{1}{2}x=\frac{1}{2}x$.

The x-intercept of a parabola y = kx² is half the x value of the point of tangency or $\frac{x}{2}$, regardless of k.

Using this knowledge, we will prove that the triangle whose vertices are the focus of a parabola, the point of tangency of a tangent line to the parabola, and the x-intercept of the tangent line is a right triangle—the vertex of the right angle being the x-intercept. This, in turn, will allow us to uncover the covert rhombus.

## The Covert Rhombus

The focus of a parabola defined by y = kx² is at (0, 1/(4k)). There hides a right triangle amongst the tangent point, focus, and the x- and y-intercepts. Once we prove that, we can prove that FPQY is a right parallelogram, or a rhombus. It's easy to see that PQ = FY. Hence FP = YQ, making this quadrilateral a parallelogram. Among all these points, there are 8 triangles, 4 of which are isosceles right triangles and 4 congruent right triangles. There are two ways to prove these findings.

Method 1

The first approach is to prove FXP is a right triangle. This can be done using the Pythagorean theorem. We can find the length of FP using distance formula and the Pythagorean theorem. If they are equal both ways, then FXP is a right triangle.

Using the distance formula: $\overline{FP}=\sqrt{{{(x-0)}^{2}}+{{\left( k{{x}^{2}}-\frac{1}{4k} \right)}^{2}}}=\sqrt{\frac{{{x}^{2}}}{2}+{{k}^{2}}{{x}^{4}}+\frac{1}{16{{k}^{2}}}}$

FX and XP can be found from the right triangles.

$\overline{FX}=\sqrt{\frac{{{x}^{2}}}{4}+\frac{1}{16{{k}^{2}}}}$
$\overline{PX}=\sqrt{\frac{{{x}^{2}}}{4}+{{k}^{2}}{{x}^{4}}}$

If $\sqrt{{{\left( \overline{PX} \right)}^{2}}+{{\left( \overline{FX} \right)}^{2}}}$ gives us the same length as using the distance formula, then FXP has to be a right triangle.

$\sqrt{{{\left( \overline{PX} \right)}^{2}}+{{\left( \overline{FX} \right)}^{2}}}=\sqrt{\frac{{{x}^{2}}}{4}+{{k}^{2}}{{x}^{4}}+\frac{{{x}^{2}}}{4}+\frac{1}{16{{k}^{2}}}}=\sqrt{\frac{{{x}^{2}}}{2}+{{k}^{2}}{{x}^{4}}+\frac{1}{16{{k}^{2}}}}$

This makes FXP a right triangle, which means FPQY is a rhombus.

Method 2

The simple method is to note that a line through Q and parallel to the x-axis is the directrix of the parabola. Hence, FP = PQ by definition of a parabola. Moreover, a ray originating from the focus reflects off the parabola perpendicular to the directrix. This makes ∠FPX = ∠XPQ. Triangle ΔFPQ is then an isosceles triangle. By nature of parallel lines, we come to the same conclusion that FPQY is a rhombus with its diagonals forming right angles.