# Putnam 1951 A6

Hey,

## Homework Statement

Putnam 1951 A6
Determine the position of a normal chord of a parabola such that it cuts off of the parabola a segment of minimum area.

## Homework Equations

Standard Form of a Parabola
$${y} = {{{a}{{x}^{2}}} + {{b}{x}} + {c}}{{\,}{\,}{\,}{\,}{\,}{\,}}{\textrm{Vertical Axis of Symmetry}}$$
$${x} = {{{d}{{y}^{2}}} + {{e}{y}} + {f}}{{\,}{\,}{\,}{\,}{\,}{\,}}{\textrm{Hortizontal Axis of Symmetry}}$$

Area Bounded Between Curves
$${A_{B}} = {\int_{a,c}^{b,d}{{\left[}{{{{f(x)}_{T}},{{f(y)}_{R}}} - {{{g(x)}_{B}},{{g(y)}_{L}}}}{\right]{{dx},{dy}}}}}{\,}{\,}{\,}{\,}{\,}{\,}{\,}{\,}{{a}{\leq}{x}{\leq}{b}}{\,}{\,}{\,}{\,}{,}{\,}{\,}{\,}{\,}{{c}{\leq}{y}{\leq}{d}}$$

## The Attempt at a Solution

From the Standard Form of a Parabola (Vertical Axis of Symmetry) let,
$${f(x)} = {{{a}{{x}^{2}}} + {{b}{x}} + {c}}$$

Since we are asked to find the position of a normal chord of a parabola, we first need to find the slope of a tangent line to a parabola for some point, ${{{P}_{1}}{{\left(}{{{x}_{1}},{{y}_{1}}}{\right)}}}$ on the parabola.
So,
$${{{f}^{\prime}}{(x)}} = {{{2}{a}{x}}+{b}}$$

Let,
$${{{f}^{\prime}}{(x)}} = {m(x)}$$

However, we need a normal line, so let,
$${p(x)} = {\frac{{-}{1}}{m(x)}}$$

$${p(x)} = {\frac{{-}{1}}{{{2}{a}{x}}+{b}}}$$

Since we need a chord normal to the parabola, we need to consider two points on the parabola: ${{{P}_{1}}{{\left(}{{{x}_{1}},{{y}_{1}}}{\right)}}}$ and ${{{P}_{2}}{{\left(}{{{x}_{2}},{{y}_{2}}}{\right)}}}$; such that a chord going through both of the points on the parabola will bound an area with only the curve of the parabola where one of the intersections will be normal to the parabola.

Let, the chord's intersection at ${{{P}_{1}}{{\left(}{{{x}_{1}},{{y}_{1}}}{\right)}}}$ be normal to the parabola.
$${p({{\left(}{{x}_{1}}{\right)}})} = {\frac{{-}{1}}{{{2}{a}{{{\left(}{{x}_{1}}{\right)}}}}+{b}}}$$

$${p{{\left(}{{x}_{1}}{\right)}}} = {\frac{{-}{1}}{{{2}{a}{{{{x}_{1}}}}}+{b}}}$$

Now consider the line,
$${y} = {{p}{x}+{c}}$$

Where ${c}$ is the y-intercept of the line.
Let ${{c} = {e}}$ to avoid confusion with ${f(x)} = {{{a}{{x}^{2}}} + {{b}{x}} + {c}}$.
$${y} = {{p}{x}+{(e)}}$$
$${y} = {{p}{x}+{e}}$$

Letting, ${{p} = {p(x)}}$ where ${{p(x)} = {p{{\left(}{{x}_{1}}{\right)}}}}$.
$${y} = {{{\left(}{{p{{\left(}{{x}_{1}}{\right)}}}}{\right)}}{x}+{e}}$$

$${y} = {{{\left(}{\frac{{-}{1}}{{{2}{a}{{{{x}_{1}}}}}+{b}}}{\right)}}{x}+{e}}$$

$${y} = {{{\frac{{-}{x}}{{{2}{a}{{{{x}_{1}}}}}+{b}}}}+{e}}$$

Now, consider the area bounded between the two curves,
$${{A}_{B}} = {{\int_{{x}_{1}}^{{x}_{2}}}{\left[}{{\left(}{{{\frac{{-}{x}}{{{2}{a}{{{{x}_{1}}}}}+{b}}}}+{e}}{\right)} - {\left(}{{{a}{{x}^{2}}} + {{b}{x}} + {c}}{\right)}}{\right]}{dx}}$$

From here is where I get kind of stuck, because once I integrate and evaluate the limits of the integral I am going to need to find a way to minimize the area. However, this is tricky since (once the integral is evaluated) the area will be given as a function of two variables, ${{A}_{B}}{{\left(}{{{x}_{1}},{{x}_{2}}}{\right)}}$.

I'm guessing the next step from there would be applying the technique of Lagrange Multipliers to find the minimum area by extracting some restrictions on the values: ${{x}_{1}}$ and ${{x}_{2}}$. Is that right? Or is there something I am missing?

Remark
I did not like the way that I started the problem from the Standard Form of a Parabola that has symmetry with respect to the vertical axis, in other words a parabola as a ${f(x)}$. I feel that had I started from the General Equation of a Conic Section where for a parabola ${{{{B}^{2}} - {{4}{A}{C}}} = {0}}$ the approach would be more rigorous, since it is the most general way to represent a parabola. Any ideas on how to approach this problem that way?

Thanks,

-PFStudent

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Hurkyl
Staff Emeritus
Gold Member
All parabolas are similar.

Hey,

All parabolas are similar.
Hmm. I don't think I follow what you mean. I understand that all parabolas are similar but just do not see the connection to helping me solve this problem. A little elaboration please?

Thanks,

-PFStudent

Last edited:
Hurkyl
Staff Emeritus