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## 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__

[tex]

{y} = {{{a}{{x}^{2}}} + {{b}{x}} + {c}}{{\,}{\,}{\,}{\,}{\,}{\,}}{\textrm{Vertical Axis of Symmetry}}

[/tex]

[tex]

{x} = {{{d}{{y}^{2}}} + {{e}{y}} + {f}}{{\,}{\,}{\,}{\,}{\,}{\,}}{\textrm{Hortizontal Axis of Symmetry}}

[/tex]

__Area Bounded Between Curves__

[tex]

{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}}

[/tex]

## The Attempt at a Solution

From the Standard Form of a Parabola (Vertical Axis of Symmetry) let,

[tex]

{f(x)} = {{{a}{{x}^{2}}} + {{b}{x}} + {c}}

[/tex]

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, [itex]{{{P}_{1}}{{\left(}{{{x}_{1}},{{y}_{1}}}{\right)}}}[/itex] on the parabola.

So,

[tex]

{{{f}^{\prime}}{(x)}} = {{{2}{a}{x}}+{b}}

[/tex]

Let,

[tex]

{{{f}^{\prime}}{(x)}} = {m(x)}

[/tex]

However, we need a normal line, so let,

[tex]

{p(x)} = {\frac{{-}{1}}{m(x)}}

[/tex]

[tex]

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

[/tex]

Since we need a chord normal to the parabola, we need to consider two points on the parabola: [itex]{{{P}_{1}}{{\left(}{{{x}_{1}},{{y}_{1}}}{\right)}}}[/itex] and [itex]{{{P}_{2}}{{\left(}{{{x}_{2}},{{y}_{2}}}{\right)}}}[/itex]; 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 [itex]{{{P}_{1}}{{\left(}{{{x}_{1}},{{y}_{1}}}{\right)}}}[/itex] be normal to the parabola.

[tex]

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

[/tex]

[tex]

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

[/tex]

Now consider the line,

[tex]

{y} = {{p}{x}+{c}}

[/tex]

Where [itex]{c}[/itex] is the y-intercept of the line.

Let [itex]{{c} = {e}}[/itex] to avoid confusion with [itex]{f(x)} = {{{a}{{x}^{2}}} + {{b}{x}} + {c}}[/itex].

[tex]

{y} = {{p}{x}+{(e)}}

[/tex]

[tex]

{y} = {{p}{x}+{e}}

[/tex]

Letting, [itex]{{p} = {p(x)}}[/itex] where [itex]{{p(x)} = {p{{\left(}{{x}_{1}}{\right)}}}}[/itex].

[tex]

{y} = {{{\left(}{{p{{\left(}{{x}_{1}}{\right)}}}}{\right)}}{x}+{e}}

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

Now, consider the area bounded between the two curves,

[tex]

{{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}}

[/tex]

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, [itex]{{A}_{B}}{{\left(}{{{x}_{1}},{{x}_{2}}}{\right)}}[/itex].

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: [itex]{{x}_{1}}[/itex] and [itex]{{x}_{2}}[/itex]. 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 [itex]{f(x)}[/itex]. I feel that had I started from the General Equation of a Conic Section where for a parabola [itex]{{{{B}^{2}} - {{4}{A}{C}}} = {0}}[/itex] 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