• Calculus > Properties of Parabolas

Properties of Parabolas

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.

October 30, 2021. I have added another well-known property that goes along with the intersection of two tangents: orthogonal tangents.

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 \(y=kx^2\). 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 for our parabola is \(y_{yi} = -kx^2\). Its derivative (which gives the slope of the tangent at x) is \(y' = 2kx\). So, a line tangent to the parabola at point p has the equation \(y = (2kp)x - kp^2\).

A tangent to a parabola \(y=kx^2\) at point (p, kp²) is defined by the equation \(y = (2kp)x - kp^2\).

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

(i) \((2ka)x-ka^2 = (2kb)x-kb^2\)

(ii) \(x(2ka-2kb) = k(a^2-b^2)\)

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

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

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

The intersection point of two tangents of a parabola \(y = kx^2\) 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 of the y values: \(\sqrt{ka^2\cdot kb^2} = \sqrt{k^2 a^2 b^2} = kab\). Here, we have to take the absolute values as the actual intersection point can be negative.

Example

In the firgure above, our points on the parabola \(y = \frac{1}{2}x^2\) are (–1.5, 1.125) and (2.2, 2.42). The intersection of the two tangents at this point is \(\left( \frac{-1.5+2.2}{2}, \frac{1}{2}\cdot (-1.5)(2.2) \right)\). This is equal to (0.35, –1.65) as calculated by Geogebra in the image above.

Remember we are using the x values of the two points to find the y value of the intersection point.

Orthogonal Tangents to Parabolas

This is a well known property that has been proved using geometry. However, we will prove this here with what we have learned above.

If two tangents of a parabola are orthogonal, then their intersection point is on the directrix of the parabola. If two tangents form a right angle at their intersection point, then the intersection point falls on the directrix of the parabola.

This is remarkable and can be proved using geometry. However, we will prove this with some simple equations of the intersection points we learned above.

Let a tangent of the parabola \(y = ax^2\) be located at \((p, ap^2)\). The slope of the tangent to this line is \(y'(p) = 2ap\). The orthogonal tangent would have to have a slope of \(-\frac{1}{2ap}\), which is the negative inverse of our tangent. Let find the point of tangency of this orthogonal tangent. The easiest methods is to use the derivative to find the x value that gives us the slope of the orthogonal:

(i) \(2ax = -\frac{1}{2ap}\)

(ii) \(x = -\frac{1}{4a^{2}p}\)

Now, we can find the y value of the point of tangency given the x value:

(iii) \(y = a\left(-\frac{1}{4a^{2}p}\right)^2 = \frac{1}{16a^{3}p^{2}}\)

Now, we know the points of tangency of orthogonals: \((p, ap^2)\) and \(\left(-\frac{1}{4a^{2}p}, \frac{1}{16a^{3}p^{2}} \right)\).

Lastly, we will find the intersection point of the two orthogonal tangents, starting with the x of the intersection. The x value is the average of the x values of the points of tangency:

(iv) \(x: \frac{p-\frac{1}{4a^2p}}{2} = \frac{4a^2 p^2 - 1}{8a^2 p}\)

The y value is the product of the x values and the coefficient, a:

(v) \(y: a(p)\left(-\frac{1}{4a^2p}\right) = -\frac{1}{4a}\)

There you have it. The intersection point of the two orthogonal tangents is \(\left(\frac{4a^2 p^2 - 1}{8a^2 p}, -\frac{1}{4a} \right)\).

No matter what the x value of the intersection point is, the y value is always \(-\frac{1}{4a}\), which is the location of the directrix of the parabola. This also means that if two tangents form a right angle at their intersection, then they fall on the directrix.

The Geogebra activity above will demonstrate what we found above. Move slider for A to move point A. You can change the coefficient of the parabola.

A secant line has also been drawn from A to B. This line passes through F, the focus of the parabola. To prove this, we simply find the equation of the secant line and set x to 0 to obtain the y value. If the y value equals 1/(4a), then it falls on the focus.

The secant line passes through the points \((p, ap^2)\) and \(\left(-\frac{1}{4a^{2}p}, \frac{1}{16a^{3}p^{2}} \right)\). The slope, m, of the secant line is:

(i) \(m = \frac{\frac{1}{16a^3p^2}-ap^2}{-\frac{1}{4a^2p}-p} = -\frac{\frac{1-16a^4p^4}{16a^3p^2}}{\frac{1+4a^2p^2}{4a^2p}}\).

(ii) \(m = -\frac{4a^2p(1-4a^2p^2)(1+4a^2p^2)}{16a^3p^2(1+4a^2p^2)}\).

(iii) \(m = -\frac{1-4a^2p^2}{4ap}\).

We can use the point \((p, ap^2)\) to write our equation of the line from the slope in (iii):

(iv) \(y - ap^2 = -\frac{1-4a^2p^2}{4ap}(x-p)\).

Now, we let x = 0:

(v) \(y = -\frac{1-4a^2p^2}{4ap}(-p) + ap^2\).

(vi) \(y = \frac{1-4a^2p^2}{4a} + ap^2\).

(vii) \(y = \frac{1-4a^2p^2 + 4a^2p^2}{4a} = \frac{1}{4a}\).

So, the y value is always 1/(4a) and the secant always passes through the focus of the parabola.

We will see this in geometric proof later, also.

I have included the geometric proof below on this page: Orthogonal Tangents Proof.

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.

Constant Area 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 light blue 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 \(y = kx^2\). The bounded area equals the area between the parabola and the first tangent from a to (a + b)/2 plus the area between the parabola and the second tangent from (a + b)/2 to b.

The first bounded area is represented by the equation:

(i) \(A_1 = \int_{a}^{(a+b)/2} kx^2 - [(2ka)x - ka^2] \text{ } dx \)

(ii) \(A_1 = \left[\frac{kx^3}{3} - (ka)x^2 + ka^{2}x\right]_{a}^{(a+b)/2} \)

(iii) \(A_1 = \left[\frac{k\left(\frac{a+b}{2}\right)^3}{3} - (ka)\left(\frac{a+b}{2}\right)^2 + ka^2\left(\frac{a+b}{2}\right)\right] - \) \(\left[\frac{ka^3}{3} - ka^3 + ka^3\right]\)

(iv) \(A_1 = \frac{k\left(\frac{a+b}{2}\right)^3}{3} - (ka)\left(\frac{a+b}{2}\right)^2 + \) \(ka^2\left(\frac{a+b}{2}\right) - \frac{ka^3}{3}\)

(v) \(A_1 = \frac{k(a+b)^3}{24} - \frac{ka(a+b)^2}{4} + \) \(\frac{ka^2(a+b)}{2} - \frac{ka^3}{3}\)

The second bounded area is represented by the equation:

(i) \(A_2 = \int_{(a+b)/2}^{b} kx^2 - [(2kb)x - kb^2] \text{ } dx\)

(ii) \(A_2 = \left[\frac{kx^3}{3} - (kb)x^2 + kb^{2}x\right]_{(a+b)/2}^{b} \)

(iii) \(A_2 = \left[\frac{kb^3}{3} - kb^3 + kb^3\right] - \)\(\left[\frac{k\left(\frac{a+b}{2}\right)^3}{3} - (kb)\left(\frac{a+b}{2}\right)^2 + kb^2\left(\frac{a+b}{2}\right)\right] \)

(iv) \(A_2 = \frac{kb^3}{3} - \frac{k(a+b)^3}{24} + \) \(\frac{kb(a+b)^2}{4} - \frac{kb^2(a+b)}{2}\)

Now, let’s add the two areas:

(i) \(A_1 + A_2 = A = \) \(\frac{k(a+b)^3}{24} - \frac{ka(a+b)^2}{4} + \) \(\frac{ka^2(a+b)}{2} - \frac{ka^3}{3} + \) \(\frac{kb^3}{3} - \frac{k(a+b)^3}{24} + \frac{kb(a+b)^2}{4} - \frac{kb^2(a+b)}{2}\)

(ii) \(A = \frac{kb^3}{3} - \frac{ka^3}{3} + \frac{kb(a+b)^2}{4} - \) \(\frac{ka(a+b)^2}{4} + \frac{ka^2(a+b)}{2} - \frac{kb^2(a+b)}{2}\)

(iii) \(A = \frac{k}{3}(b-a)(a^2+ab+b^2) + \) \(\frac{k(a+b)^2}{4}(b-a) - \frac{k(a+b)^2}{2}(b-a)\)

(iv) \(A = \frac{k(b-a)}{12}\left[4a^2+4ab+4b^2 - 3a^2-6ab-3b^2\right]\)

(v) \(A = \frac{k(b-a)}{12}(a^2-2ab+b^2) =\) \( \frac{k(b-a)^3}{12}\)

(vi) \(A = \frac{k(b-a)^3}{12}\)

If we let d be the distance from a to b, then the area is equal to \(A = \frac{kd^3}{12}\).

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

Therefore, regardless of where a and b are located on the parabola, if the horizontal distance between them is constant, the area enclosed by the tangents and the parabola is constant.

Fun fact: a vertical line through the intersection of the two tangents divides the area in half. In the image above, you can see that the two halves have an area of approximately 1.056 as reported by Geogebra. The total area would be 2.112. Using our formula we get \(A = \frac{kd^3}{12} = \frac{(0.5)(2.2 - (-1.5))^3}{12} = \) \(\frac{(0.5)(3.7)^3}{12} \approx 2.11054\)

Constant Area 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 intergration is fairly simple for this setup.

First, we find the equation of the secant line that passes through the points named above:

(i) \(y-ka^2 = \frac{kb^2-ka^2}{b-a}(x-a)\)

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

(iii) \(y=k(a+b)x-ka(a+b)+ka^2\)

(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:

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

(ii) \(A=\left[ \frac{k(a+b)b^2}{2}-kab^2-\frac{kb^3}{3} \right] - \) \( \left[ \frac{k(a+b)a^2}{2}-ka^2b-\frac{ka^3}{3} \right]\)

Skip a few steps in between the rearranging and factoring:

(iii) \(A= \frac{1}{2}k(a+b)^2(b-a) - kab(b-a) - \) \(\frac{k}{3}(b-a)(a^2+ab+b^2)\)

(vi) \(A=\frac{1}{6}k(b-a)[3(a+b)^2 - 6ab - 2(a^2+ab+b^2)]\)

(v) \(A=\frac{1}{6}k(b-a)(a-b)^2\)

Since (a – b)² is the same as (b – a)², our last expression reduces nicely. If we let d be the distance between a and b, then we have the following:

The area, A, bound by a secant passing through (a, ka²) and (b, kb²) and a parabola is constant and is only dependent on the horizontal distance, d, between the two points. The area is given by the formula: \(\frac{kd^3}{6}\)!

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. We found this area above already.

The total area of the triangle formed by the secan and the two tangents at these points is \(\frac{kd^3}{12} + \frac{kd^3}{6} = \frac{kd^3}{4}\).

Above is a Geogebra activity that you can play with. You can adjust the horizontal distance, Δd between points A and B and see the effects on the enclosed area. Drag points A and B and notice the area is constant. You can also change the constant a to change the shape of the parabola.

Example

Let’s do the messy calculus with an example and then use our formula to verify our result above. Consider a secant passing through (–1, 1/4) and (2, 1) on the parabola \(y=\frac{1}{4}x^2\). The slope of this secant is \(m = \frac{1-(1/4)}{2-(-1)} = \frac{1}{4}\). Using the point (2, 1), the equation of this line is \(y-1=\frac{1}{4}(x-2)\) or \(y = \frac{1}{4}x+\frac{1}{2}\). Now, let’s integrate with the bounds –1 to 2 between this line and the parabola:

(i) \(A = \int_{-1}^{2} \frac{1}{4}x +\frac{1}{2}-\frac{1}{4}x^2\text{ }dx\)

(ii) \(A = \left[ \frac{1}{8}x^2 + \frac{1}{2}x - \frac{1}{12}x^3 \right]_{-1}^{2}\)

(iii) \(A = \left( \frac{4}{8} + \frac{2}{2} - \frac{8}{12} \right) - \left( \frac{1}{8} - \frac{1}{2} + \frac{1}{12} \right) = \) \(\frac{5}{6} + \frac{7}{24} = \frac{9}{8}\)

Using our formula \(A = \frac{kd^3}{6}\) from above, k = 1/4 and d = 2 – (–1) = 3. So the area is \(A = \frac{\frac{1}{4}(3^3)}{6} = \frac{27}{24} = \frac{9}{8}\).

The area is the same using both methods.

The XY-Intercepts

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

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

The x-intercept of a parabola y = kx² is half the x value of the point of tangency or \(y_{xi}(x) = \frac{x}{2}\), regardless of k. The y-intercept is the same vertical distance as the point of tangency or \(y_{yi}(x) = -kx^2\).

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. We know that PQ = FY because we stated this fact about y-intercepts in the previous section. Hence FP has to equal YQ because FY and PQ are parallel and equal. That makes quadrilateral FPQY a parallelogram. There are two ways to prove this.

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 or the Pythagorean theorem. If they are equal both ways, then FXP is a right triangle. Using the distance formula, we have:

(i) \(\overline{FP}=\sqrt{(x-0)^2+\left( kx^2-\frac{1}{4k} \right)^2}\)

(ii) \(\overline{FP} = \sqrt{\frac{x^2}{2}+k^{2}x^4+\frac{1}{16k^2}} =\) \(kx^2+\frac{1}{4k}\)

FX and XP can be found from the right triangles.

(iii) \(\overline{FX}=\sqrt{\frac{x^2}{4}+\frac{1}{16k^2}}\)

(iv) \(\overline{PX}=\sqrt{\frac{x^2}{4}+k^{2}x^4}\)

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.

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

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, showin in green. 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.

Geometric Proof of Orthogonal Tangents and Directrix

Earlier on this page, I gave the proof that the intersection of orthogonal tangents of a parabola fall on the directrix of the parabola by using the intersection points. Here, I will show a geometric proof. We will refer to this image.

First, we have a parabola with tangents at A and B, which form a right angle at C because they are orthogonal. The focus of the parabola is at F. AE has been drawn where E is on the directrix and BD has been drawn where D is on the directrix.

We need to prove that FO is equal to OH since the directrix is the same distance as FO from the x-axis.

First, note that AF = AE by the definition of parabolas. Triangles AGE and AGF are congruent right triangles, which was covered in the rhombus section. That means FG = GE. Also DC = CE because the intersection of tangents is the midpoint of the horizontal distance, covered earlier.

If we draw a circle with diameter of DE, then AE will be perpendicular to DE because it forms a right angle there. And since AE = AF, AF is also tangent to this circle.

That makes FC the circumradius of the circle and HFE a right triangle. GOF is similar to EFH and EFH is twice as big. That means FO = OH.

There are actually many similar triangles in this image. We can also note the focus falls on the same line as AB. That leads us to the following corollary.

If a secant of a parabola passes through the focus of the parabola, then the tangents at the intersection of the secants to the parabola will be orthogonal.

Last Word

These are just a few properties of parabolas. I have found some more properties that are presented on The Parabola Revisited page. Another pearl is that when the parabola rolls on the x-axis, its focus traces a catenary, which is the hyperbolic cosine function. I have the proof given here with the Geogebra interactions: Proof Focus of Parabolic Traces a Catenary.

I have discussed parabolas also in the Conic Sections page also and given their equations in terms of eccentricty.

There are probably more undiscovered properties as well.

Anirdesh.com