Lesson Another Cool Analytic Geometry Problem

MarkFL

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Consider the following diagram:

tml_extremenormal.png


In blue, is a parabola of the form:

\(\displaystyle y=ax^2\) where \(0<x\).

Let us define a normal line as a line that is orthogonal to the parabola in the first quadrant. Five normal lines are shown in the figures above. The significance of the solid line in black will be explained later.

For a while, the \(x\)-coordinate of the second quadrant intersection of a normal line with the parabola gets smaller as the x-coordinate of the first quadrant intersection gets smaller. But eventually a normal line's second quadrant intersection gets as small as it can get. When this happens, we have the extreme normal line, and this is the solid black line above. Once the normal lines pass the extreme normal line, the \(x\)-coordinate of the second quadrant intersections with the parabola start to increase.

A normal line pair is two normal lines that have the same quadrant II point of intersection with the parabola. The following diagram shows two such pairs:

tml_extremenormalpairs.png


In any given pair, where there are two distinct lines, one will have its first quadrant intersection with the parabola above that of the extreme normal line, and the other will be below.

Questions:

(a) Find the normal line pair associated with a particular second quadrant intersection with the parabola.

(b) Hence, or otherwise, find the equation of the extreme normal line.

(c) Find the normal line that traps the smallest area between it and the parabola.

Answers:

Let the first quadrant intersection between a normal line and the parabola be \(\left(p,ap^2 \right)\) and the second quadrant intersection be \(\left(q,aq^2 \right)\).

Because the normal line as defined is perpendicular to the parabola at the first quadrant intersection, we may express the slope \(m\) of the normal line as:

\(\displaystyle m=-\frac{1}{f'(p)}=-\frac{1}{2ap}\)

Using this slope, the first quadrant point of intersection, and the point-slope formula, we find then that the normal line is given by:

\(\displaystyle y-ap^2=-\frac{1}{2ap}(x-p)\)

Arranging this in slope-intercept form, we obtain the general normal line:

\(\displaystyle y=-\frac{1}{2ap}x+\frac{2a^2p^2+1}{2a}\tag{1}\)

(a) To find the pair of normal lines associated with a particular second quadrant intersection, we may equate two expressions for the slope of the normal line:

\(\displaystyle m=\frac{aq^2-ap^2}{q-p}=-\frac{1}{2ap}\)

\(\displaystyle a(q+p)=-\frac{1}{2ap}\)

Writing this equation in standard quadratic form in \(p\), there results:

\(\displaystyle p^2+qp+\frac{1}{2a^2}=0\)

Application of the quadratic formula yields:

\(\displaystyle p=\frac{-aq\pm\sqrt{a^2q^2-2}}{2a}\)

Using these values for \(p\), we obtain the pair of normal lines:

\(\displaystyle y=-\frac{1}{2a\left(\frac{-aq\pm\sqrt{a^2q^2-2}}{2a} \right)}x+\frac{2a^2\left(\frac{-aq\pm\sqrt{a^2q^2-2}}{2a}\right)^2+1}{2a}\)

Simplifying, we obtain:

\(\displaystyle y=\frac{1}{aq\pm\sqrt{a^2q^2-2}}x+\frac{q\left(aq\pm\sqrt{a^2q^2-2}\right)}{2}\)

(b) To find the extreme normal line, we need only equate the discriminant to zero in the solution for \(p\) for a given normal line pair, that is to take the pair which are the same line:

\(\displaystyle p=-\frac{q}{2}\)

Now, using the equation:

\(\displaystyle p^2+qp+\frac{1}{2a^2}=0\)

and solving for \(q\), we get:

\(\displaystyle q=-\frac{2a^2p^2+1}{2a^2p}\)

Hence:

\(\displaystyle p=\frac{2a^2p^2+1}{4a^2p}\)

Solving this for \(p\), we find:

\(\displaystyle p=\frac{1}{a\sqrt{2}}\)

Substituting this value of \(p\) into (1), we obtain the extreme normal line:

\(\displaystyle y=-\frac{1}{\sqrt{2}}x+\frac{1}{a}\)

We may also use the calculus to determine the extreme normal line by minimizing $q$. Recall we found:

\(\displaystyle p^2+qp+\frac{1}{2a^2}=0\)

Implicitly differentiating with respect to \(p\), we find:

\(\displaystyle 2p+q+p\frac{dq}{dp}=0\)

\(\displaystyle \frac{dq}{dp}=-2-\frac{q}{p}\)

Equating this to zero, we find:

\(\displaystyle p=-\frac{q}{2}\)

Finding the extreme normal line now follows the same method as above.

(c) To find the normal line that traps the minimal area \(A\) between it and the parabola, we may begin by expressing this area as:

\(\displaystyle A=\int_q^p -\frac{1}{2ap}x+\frac{2a^2p^2+1}{2a}-ax^2\,dx\)

The number and nature of the extrema will not depend upon the value of \(a\), so first let's let \(a=1\):

\(\displaystyle A=\int_{-\left(p+\frac{1}{2p} \right)}^p -\frac{1}{2p}x+ \frac{2p^2+1}{2}-x^2\,dx\)

Completing the square on the integrand, we have:

\(\displaystyle A=\int_{-\left(p+\frac{1}{2p}\right)}^p\left(p+\frac{1}{4p}\right)^2-\left(x+\frac{1}{4p}\right)^2\,dx\)

Using the substitution:

\(\displaystyle u=x+\frac{1}{4p}\,\therefore\,du=dx\)

and the even function rule, we may write:

\(\displaystyle A=2\int_{0}^{p+\frac{1}{4p}} \left(p+\frac{1}{4p} \right)^2-u^2\,du\)

Applying the FTOC, we find:

\(\displaystyle A=\frac{4}{3}\left(p+\frac{1}{4p}\right)^3\)

Differentiating with respect to \(p\), we find:

\(\displaystyle \frac{dA}{dp}=4\left(p+\frac{1}{4p}\right)^2\left(1-\frac{1}{4p^2}\right)\)

Equating this to zero, we find the only real root comes from:

\(\displaystyle 4p^2-1=(2p+1)(2p-1)=0\)

Taking the positive root, our one critical value is:

\(\displaystyle p=\frac{1}{2}\)

The first derivative test easily confirms that this a minimum.

And so, using (1), the normal line that traps the minimal area is:

\(\displaystyle y=-x+\frac{3}{4}\)

Using the same reasoning, the normal line trapping the minimal area for an arbitrary \(a\) is when:

\(\displaystyle p=\frac{1}{2a}\)

Using (1) again, the normal line that traps the minimal area is:

\(\displaystyle y=-x+\frac{3}{4a}\)

And so we find the minimal trapped area to be:

\(\displaystyle A_{\min}=\frac{4}{3a^2}\)
 
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harpazo

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OMG! Outstanding!!
 

MarkFL

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Here is a link to a live Desmos graph with sliders to explore the normal lines.

Desmos graph
 
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MarkFL

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And here is a live graph to explore normal line pairs.

Desmos graph
 
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Reactions: anemone and harpazo