Chapter 4, Problem 15P

(a)

To determine

To show: If $P\left(a,a^2\right)$ is any point on the parabola $y=x^2$ but not at the origin and Q is any point on the parabola where the normal line at P intersects the parabola, then the y- coordinate of Q is smallest when $a=\frac{1}{\sqrt{2}}$

Explanation of Solution

The equation of the parabola is given as $y=x^2$.

On differentiating with respect to x, $\frac{dy}{dx}=2x$.

Thus, the slope of the tangent is, $\frac{dy}{dx}=2x$.

So, the slope of the normal is, $\frac{-1}{2x}$.

At the point $\left(a,a^2\right)$, the slope of tangent is $2a$ and the slope of normal is $\frac{-1}{2a}$.

By slope point form, the equation of normal is $y-a^2=\frac{-1}{2a}\left(x-a\right)$.

Use $y=x^2$ in the equation of normal.

$\begin{array}{l}y-a^2=\frac{-1}{2a}\left(x-a\right)\\ x^2-a^2=\frac{-1}{2a}\left(x-a\right)\\ {\left(x+\frac{1}{4a}\right)}^2={\left(a+\frac{1}{4a}\right)}^2\\ x=a\text{ or }\left(-a-\frac{1}{2a}\right)\end{array}$

One is the x-coordinate of P say $x_P=a$ and other is x-coordinate of Q say $x_Q=-\left(a+\frac{1}{2a}\right)$.

Using the value of $x_Q$ in the equation of parabola, write $y_Q={\left(a+\frac{1}{2a}\right)}^2$.

Find the derivative of $y_Q$ with respect to a as follows.

$\frac{d{y}_Q}{da}=2\left(a+\frac{1}{2a}\right)\left(1+\frac{-1}{2a^2}\right)$

Set $\frac{d{y}_Q}{da}=0$,

$\begin{array}{l}\frac{d{y}_Q}{da}=0\\ 2\left(a+\frac{1}{2a}\right)\left(1+\frac{-1}{2a^2}\right)=0\\ \left(2a+\frac{1}{a}\right)\left(2+\frac{-1}{a^2}\right)=0\\ \left(\frac{2a^2+1}{a}\right)\left(\frac{2a^2-1}{a^2}\right)=0\\ 2a^2+1=0\text{ or 2}{a}^2-1=0\end{array}$

On simplification,

$\begin{array}{l}2a^2+1=0\text{ or 2}{a}^2-1=0\\ a=\sqrt{\frac{-1}{2}}\text{ or }a=\sqrt{\frac{1}{2}}\\ a=\sqrt{\frac{-1}{2}}\text{ or }a=\frac{1}{\sqrt{2}}\end{array}$

Since a is a real number, the value of a is $a=\frac{1}{\sqrt{2}}$.

Compute the second derivative of $y_Q$ with respect to a as follows.

$\begin{array}{c}\frac{d^2{y}_Q}{d{a}^2}=\frac{d}{da}\left(2\left(a+\frac{1}{2a}\right)\left(1+\frac{-1}{2a^2}\right)\right)\\ =2\left\{\left(a+\frac{1}{2a}\right)\frac{d}{da}\left(1+\frac{-1}{2a^2}\right)+\left(1+\frac{-1}{2a^2}\right)\frac{d}{da}\left(a+\frac{1}{2a}\right)\right\}\left[\text{By product rule}\right]\\ =2\left\{\left(a+\frac{1}{2a}\right)\left(\frac{1}{a^3}\right)+\left(1+\frac{-1}{2a^2}\right)\left(1-\frac{1}{2a^2}\right)\right\}\\ =\frac{4a^4+3}{2a^4}\end{array}$

Clearly for any value of a, $\frac{d^2{y}_Q}{d{a}^2}>0$.

That is, the value of y coordinate is minimum when $a=\frac{1}{\sqrt{2}}$.

Hence the proof.

(b)

To determine

To show: If $P\left(a,a^2\right)$ is any point on the parabola $y=x^2$ and Q any point where normal line P intersects the parabola then the line segment PQ has the shortest possible length when $a=\frac{1}{\sqrt{2}}$.

Explanation of Solution

Use the information deduced in part (a).

If $P\left(a,a^2\right)$ is the given point, then the point Q is given by $Q\left(-a-\frac{1}{2a},{\left(-a-\frac{1}{2a}\right)}^2\right)$.

Let the distance between these two points is $D_{PQ}$. Then,

$\begin{array}{c}{D}_{PQ}{}^2={\left(\sqrt{{\left(2a+\frac{1}{2a}\right)}^2+{\left(a^2-{\left(a+\frac{1}{2a}\right)}^2\right)}^2}\right)}^2\\ =4a^2+2+\frac{1}{4a^2}+{\left(a^2-\left(a^2+1+\frac{1}{4a^2}\right)\right)}^2\\ =4a^2+2+\frac{1}{4a^2}+1+\frac{1}{2a^2}+\frac{1}{16a^4}\end{array}$

Differentiate the above equation with respect to a.

$2D_{PQ}\frac{d{D}_{PQ}}{da}=8a-\frac{1}{2a^3}-\frac{1}{a^3}-\frac{1}{4a^5}$

Equate the above equation to 0.

$\begin{array}{c}8a-\frac{1}{2a^3}-\frac{1}{a^3}-\frac{1}{4a^5}=0\\ \frac{32a^6-2a^2-4a^2-1}{4a^5}=0\\ 32a^6-6a^2-1=0\end{array}$

On simplifying the above equation, the only possible value of a is $a=\frac{1}{\sqrt{2}}$.

Compute the second derivative of $D_{PQ}$ with respect to a as follows.

$\frac{d^2{D}_{PQ}}{d{a}^2}=8+\frac{9}{2a^4}+\frac{5}{4a^6}$

As $\frac{d^2{D}_{PQ}}{d{a}^2}>0$, this is the minimum distance between P and Q.

That is the segment PQ has the shortest possible length when $a=\frac{1}{\sqrt{2}}$.

Hence the proof.