A Fixed-Length Chord Inside a Parabola

The college entrance examination (Gaokao) results are out, and they are not very ideal; I cannot enter a very ideal university. However, no matter where I go, I will continue my scientific dream, devoted to mathematical and physical research. Yesterday, the college preference submission was also completed, so the matters of the college entrance examination have come to an end for now, and I am just waiting for the notice.

The following few articles may explore some interesting locus problems related to conic sections, which were basically completed within the two weeks before the college entrance examination. Let’s look at the simplest one: there is a chord of fixed length a inside the parabola y=x^2. Find the locus of the midpoint of the chord and explore the position of the lowest point of the locus.

We can imagine a corresponding physical model: a large bowl in the shape of a paraboloid of revolution, with a chopstick sliding down inside. Find the locus of the midpoint. What does the lowest point of the midpoint mean? According to the definition of gravitational potential energy, the lowest midpoint means the minimum potential energy. The “principle of minimum potential energy” tells us that minimum potential energy implies equilibrium. It turns out that the lowest midpoint actually corresponds to the stable position of this chopstick.

First, let’s find the locus. As shown in the figure, let A=(x_1,y_1) and B=(x_2,y_2). Combined with y=x^2, we can list the relationship: (x_1-x_2)^2+(x_1^2-x_2^2)^2=a^2

The expression for the midpoint (x, y) is: x=\frac{x_1+x_2}{2}, \quad y=\frac{x_1^2+x_2^2}{2}

This form of expression requires us to try to bring the distance relationship closer to x_1+x_2 and x_1^2+x_2^2. The process is: \begin{aligned} (x_1-x_2)^2[1+(x_1+x_2)^2] &= a^2 \\ [2(x_1^2+x_2^2)-(x_1+x_2)^2][1+(x_1+x_2)^2] &= a^2 \end{aligned}

Substituting the midpoint coordinates, the locus equation is: 4(y-x^2)(1+4x^2)=a^2 Or written as: 4y=\frac{a^2}{1+4x^2}+4x^2

Now let’s explore the position of the lowest point of the locus. That is, find the minimum value of y. From the fundamental inequality (AM-GM), we can get: 4y=\frac{a^2}{1+4x^2}+(1+4x^2)-1 \geq 2\sqrt{\frac{a^2}{1+4x^2}\cdot (1+4x^2)}-1=2a-1

That is, y_{\min}=\frac{2a-1}{4}. However, there is a constraint: the condition for the equality to hold is \frac{a^2}{1+4x^2}=1+4x^2, which requires a \geq 1.

When a \leq 1, it is easy to find that the minimum value y_{\min}=\frac{a^2}{4} is reached when x=0.

The images corresponding to different values of a are as follows:

As mentioned before, the lowest point is the equilibrium position of that “chopstick”. However, force analysis tells us that when AB is parallel to the x-axis (horizontal position), the chopstick is also in force equilibrium and should also belong to the equilibrium points. Why is the point we found not this one? In fact, a system often has multiple equilibrium positions, but not every equilibrium position is stable. A stable equilibrium position will only oscillate slightly around the equilibrium point after a slight disturbance, while an unstable equilibrium point will deviate far from its original state after a slight disturbance. For example, if we gently shake this bowl, how will this chopstick swing? Swinging slightly at the original position means it is stable; moving away from the original position to another position means it is unstable—even though they are both force equilibrium points.

For a locus where the parabola is in the form y=kx^2, it can be converted into the form in this article through substitution, so I will not repeat it. An easy question to think of is: if the curve in this article is changed to an ellipse, what would the locus of the midpoint of the fixed-length chord be? This will be discussed in the next article.

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