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The maximum possible area bounded by the parabola $y = x^2 + x + 10$ and a chord of the parabola of length $1$ is?

$(y-39/4)=(x+1/2)^2$, Vertex: $(-1/2, 3/4)$

How do I find the equation of the chord whose length is $1$?

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  • $\begingroup$ Did you mean vertex $(-1/2, 39/4)$? $\endgroup$ – N. F. Taussig Aug 3 at 22:38
  • $\begingroup$ Note a chord is line segment with each end point on the parabola, so these end points, call them $(x_1,y_1)$ and $(x_2,y_2)$, must have $y_i = x_i^2 + x_i + 10$ for $i=1,2$. For the chord length to be $1$ requires that $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = 1 \implies (x_1 - x_2)^2 + (y_1 - y_2)^2 = 1$. $\endgroup$ – John Omielan Aug 3 at 23:50
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Let the equation of the secant line be $y=mx+b$. Combining this with the equation of the parabola produces the quadratic equation $$x^2+(1-m)x+(10-b)=0.$$ To reduce clutter, I’ll denote the discriminant of this equation by $\Delta = (1-m)^2-4(10-b)$. The solutions to this equation are, of course, $$x = \frac12(m-1)\pm\frac12\sqrt\Delta$$ and substituting this into the parabola’s equation results in $$y=\frac14(m^2+\Delta+39)\pm\frac12m\sqrt\Delta.$$ The square of the length of the corresponding chord is therefore simply $(1+m^2)\Delta$. Setting this equal to $1$ and solving for $b$ yields $$b=10-\frac14(1-m)^2+{1 \over 4(1+m^2)},$$ so you now have the equations of a family of secants with chord length equal to $1$ parameterized by slope, but you really only need $\Delta$ and $b$ to solve the problem.

There’s a property of parabolas that’s handy for attacking this problem: The area bounded by a parabola and its chord is equal to two-thirds of the area of the bounding paralellogram (see here for details). The problem thus becomes one of finding the value of the slope $m$ that maximizes the area of this paralellogram, which is a lot easier to compute. In fact, since the chord length is fixed at $1$, the problem reduces even further to finding the value of $m$ for which the distance between the chord and the tangent parallel to it is maximized. By symmetry, that’s likely to occur when the tangent is at the vertex, i.e., when $m=0$.

Either by differentiating or by using the fact that midpoints of parallel chords lie on a line parallel to the parabola’s axis, you can find that the tangent to the parabola has a slope of $m$ when $x=\frac12(m-1)$. Substituting the value of $b$ computed above into the expression for $\Delta$ simplifies it to $\Delta = \frac1{1+m^2}$. The distance between the secant and this tangent is equal to the difference between the $y$-coordinates of the midpoint of the chord and the point of tangency†, divided by the normalizing factor of $\sqrt{1+m^2}$, which simplifies to $$\frac\Delta{4\sqrt{1+m^2}} = \frac1{4(1+m^2)^{3/2}}.$$ Therefore, the area of the parabolic segment is $\frac1{6(1+m^2)^{3/2}}$, which obviously has its maximum at $m=0$, as suspected.


† Alternatively, find the $y$-intercept of the tangent by solving $\Delta=0$ for $b$, and subtract this from the value of $b$ computed above for the secant.

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Such a chord would intersection the parabola at points $(x_1, y_1)$ and $(x_2, y_2)$, where $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = 1$. Notice that if you square this expression, you get $(y_2 - y_1)^{2} = 1 - (x_2 - x_1)^2$. Here, you'd have $y_2 = x^{2}_{2} + x_2 + 10$ and $y_1 = x^{2}_{1} + x_1 + 10$.

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  • $\begingroup$ If you square $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = 1$, you get $(x_2 - x_1)^2 + (y_2 - y_1)^2 = 1$. However, in general, and also in this case, I don't see how you can then get $y_2 - y_1 = -(x_2 - x_1)$. If this were true, then you'd get $|y_2 - y_1| = |x_2 - x_1| = \frac{1}{\sqrt{2}}$. $\endgroup$ – John Omielan Aug 4 at 0:33
  • $\begingroup$ I didn't notice that mistake. Thank you for pointing that out. $\endgroup$ – Amy Ngo Aug 4 at 0:34
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I consider the more general parabola

$y(x) = x^2 + ax + b $ with the points being $(x_i, y_i)_{i=1}^2 $.

I am close to a solution, but am pooping out so am leaving my answer incomplete.

One surprising result I find is that the area between the chord and the parabola is $\dfrac{(x_2-x_1)^3}{6} $.

The length of the chord is

$\begin{array}\\ L^2 &=(x_2-x_1)^2+(y_2-y_1)^2\\ &=(x_2-x_1)^2+((x_2^2 + ax_2 + b)-(x_1^2 + ax_1 + b))^2\\ &=(x_2-x_1)^2+(x_2^2-x_1^2 + a(x_2-x_1))^2\\ &=(x_2-x_1)^2+((x_2-x_1)(x_2+x_1) + a(x_2-x_1))^2\\ &=(x_2-x_1)^2+(x_2-x_1)^2((x_2+x_1) + a)^2\\ &=(x_2-x_1)^2(1+((x_2+x_1) + a)^2)\\ \end{array} $

If $x_2-x_1 = d$, $L^2 =d^2(1+((2x_1+d) + a)^2) $.

The equation of the chord is

$\begin{array}\\ \dfrac{y-y_1}{x-x_1} &=\dfrac{y_2-y_1}{x_2-x_1}\\ &=\dfrac{(x_2^2+ax_2+b)-(x_1^2+ax_1+b)}{x_2-x_1}\\ &=\dfrac{x_2^2-x_1^2+a(x_2-x_1)}{x_2-x_1}\\ &=x_2+x_1+a\\ &=u\\ \end{array} $

where $u = x_2+x_1+a$,

or

$\begin{array}\\ y &=y_1+(x-x_1)u\\ &=ux+x_1^2+ax_1+b-ux_1\\ &=ux+x_1^2+ax_1+b-(x_2+x_1+a)x_1\\ &=ux+b-x_2x_1\\ \end{array} $

The area under the chord is

$\begin{array}\\ A_c &=\int_{x_1}^{x_2} (ux+y_1-ux_1)dx\\ &=(\dfrac{ux^2}{2}+y_1x-ux_1x)|_{x_1}^{x_2}\\ &=\dfrac{u(x_2^2-x_1^2)}{2}+y_1(x_2-x_1)-ux_1(x_2-x_1)\\ &=(x_2-x_1)(\dfrac{u(x_2+x_1)}{2}-ux_1+y_1)\\ &=(x_2-x_1)(\dfrac{u(x_2-x_1)}{2}+y_1)\\ &=(x_2-x_1)(\dfrac{(x_2+x_1+a)(x_2-x_1)}{2}+y_1)\\ &=(x_2-x_1)(\dfrac{x_2^2-x_1^2+a(x_2-x_1)}{2}+x_1^2+ax_1+b)\\ &=(x_2-x_1)(\dfrac{x_2^2+x_1^2+a(x_2+x_1)}{2}+b)\\ \end{array} $

The area under the parabola is

$\begin{array}\\ A_p &=\int_{x_1}^{x_2} (x^2+ax+b)dx\\ &=(\dfrac{x^3}{3}+a\dfrac{x^2}{2}+bx)|_{x_1}^{x_2}\\ &=\dfrac{x_2^3-x_1^3}{3}+a\dfrac{x_2^2-x_1^2}{2}+b(x_2-x_1))\\ &=(x_2-x_1)\left(\dfrac{x_2^2+x_2x_1+x_1^2}{3}+a\dfrac{x_2+x_1}{2}+b\right)\\ \end{array} $

The difference is, noting that the $a$ and $b$ terms cancel out,

$\begin{array}\\ A_{cp} &=A_c-A_p\\ &=(x_2-x_1)\left(\dfrac{x_2^2+x_1^2}{2}-\dfrac{x_2^2+x_2x_1+x_1^2}{3}\right)\\ &=(x_2-x_1)\left(\dfrac{x_2^2+x_1^2-2x_2x_1}{6}\right)\\ &=\dfrac{(x_2-x_1)^3}{6}\\ \end{array} $

I find this very surprising.

This is the kind of result that, if it is really true, probably has a much simpler proof.

Now to incorporate the length restriction.

We have $L^2 =(x_2-x_1)^2(1+((x_2+x_1) + a)^2) $.

So we want to minimize $(x_2-x_1)^3$ subject to $L^2 =(x_2-x_1)^2(1+((x_2+x_1) + a)^2) $.

The following discussion is taken from https://en.wikipedia.org/wiki/Lagrange_multiplier

The Lagrangian is $h(x_1, x_2, \lambda) =(x_2-x_1)^3 -\lambda((x_2-x_1)^2(1+((x_2+x_1) + a)^2)-L^2) $ or, using simplified variables ($x_1 \to x, x_2 \to y, \lambda \to z$), $h(x, y, z) =(y-x)^3 -z((y-x)^2(1+((y+x) + a)^2)-L^2) =f(x, y) -zg(x, y) $.

So we want to solve

$g(x, y) = 0, \nabla_{x, y}f(x, y) =z\nabla_{x, y}g(x, y) $ where $\nabla_{x, y}f(x, y) =\left(\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}\right) $.

I think this is close to the end, but it's late and I'm tired.

So, once again, I am leaving an incomplete answer in the hope that someone else can finish it.

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    $\begingroup$ Not entirely surprising. The area bounded by a parabola and a chord is equal to two-thirds the area of the bounding paralellogram. $\endgroup$ – amd Aug 4 at 8:03
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The parabola has the same shape as $y=x^2$. The chord length of $1$ parallel to the $x$-axis connects points $\big(\pm\frac 12, \frac14\big)$. By symmetry this chord gives the largest area.

See Desmos illustration here. For simplicity, in the illustration, the chord is fixed as the segment between $(0,0)$ and $(1,0)$, and the parabola rotates.

enter image description here enter image description here enter image description here enter image description here

The area bounded by the parabola and the chord is given by $$2\int_0^{\frac 12}\frac 14-x^2 \;\;dx=\frac 16$$

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