Chapter 7, Problem 14P

To determine

To find : all the curves with the property that if the normal line is drawn at any point $P$ on the curve

Answer to Problem 14P

The curve is $x^2+\frac{y^2}{2}=C$

Explanation of Solution

Given information : The question is

The part of the normal line between $P$ and the x-axis is bisected by the y-axis

Calculation :

Assume that x-intercept of normal is $\left(x_i,0\right)$

Assume that y-intercept of normal is $\left(0,y_i\right)$

Assume that the point $P$ is $\left(a,f\left(a\right)\right)$

Since the y-intercept is midpoint,

  $\begin{array}{l}\left(0,y_i\right)=\left(\frac{x_i+a}{2},\frac{0+f\left(a\right)}{2}\right)\\ 0=\frac{x_i+a}{2}\Rightarrow 0=x_i+a\Rightarrow x_i=-a\\ y_i=\frac{0+f\left(a\right)}{2}\Rightarrow y_i=\frac{f\left(a\right)}{2}\end{array}$

We know that the slope of normal is $-\frac{1}{f\left(a\right)}$

Slope of the line joining the two intercepts is also the slope of the normal

Therefore,

  $\begin{array}{l}\frac{y_i-0}{0-x_i}=-\frac{1}{f\left(a\right)}\\ \frac{\frac{f\left(a\right)}{2}-0}{0-\left(-a\right)}=-\frac{1}{f\left(a\right)}\\ \frac{f\left(a\right)}{2}=-\frac{1}{f\left(a\right)}\\ \frac{f\left(x\right)}{2x}=-\frac{1}{f\left(a\right)}\\ \frac{y}{2x}=-\frac{1}{f\left(a\right)}\\ \frac{2x}{y}=-y\text{'}\\ -\frac{2x}{y}=\frac{dy}{dx}\\ -2xdx=ydy\\ -{\displaystyle \int 2xdx={\displaystyle \int ydy}}\\ -x^2+C=\frac{y^2}{2}\\ C=x^2+\frac{y^2}{2}\\ x^2+\frac{y^2}{2}=C\end{array}$