Chapter 3.2, Problem 58E

To determine

To show : the three normal from point $\left(a,0\right)$ to the parabola $x=y^2$ , then $a$ must be greater than $1/2$ .

Explanation of Solution

Given information :

Curve $x=y^2$ and point $\left(a,0\right)$ .

Formula needed :

Slope of tangent line is $m={\frac{dy}{dx}}_{\left(a,b\right)}$ and slope of normal line is $-\frac{1}{m}$ .

Differentiate $x=y^2$ with respect to $x$ ,

  $\begin{array}{c}\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(y^2\right)\\ 1=2y\cdot \frac{d}{dx}\left(y\right)\\ \frac{1}{2y}=\frac{dy}{dx}\\ \frac{dy}{dx}=\frac{1}{2y}\end{array}$

So, the slope of normal is $-2y$ .

As the normal line passes through points $\left(a,0\right)$ and $\left(y^2,y\right)$ . So, the slope of normal line is

  $\begin{array}{c}{m}_n=\frac{y-0}{y^2-a}\\ =\frac{y}{y^2-a}\end{array}$

Equate $\frac{y}{y^2-a}$ and $-2y$ equal to each other.

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

So,

  $\begin{array}{c}1+2y^2=2a\\ \frac{1+2y^2}{2}=a\\ a\ge \frac{1}{2}\end{array}$

The two normal are perpendicular to each other if the product of their slopes is negative 1.

Consider two points on the curve are $\left(y_1^2,y_1\right)$ and $\left(y_2^2,y_2\right)$ . Their slopes are $-2y_1$ and $-2y_2$ .

As both points are symmetric to x-axis, $y_2=-y_1$ .

The product of slopes is $-2y_1\cdot \left(-2y_2\right)=-1$ .

  $\begin{array}{c}-2y_1\cdot \left(-2\left(-y_1\right)\right)=-1\\ -2y_1\cdot 2y_1=-1\\ y_1{}^2={\left(\frac{1}{2}\right)}^2\\ y_1=\pm \frac{1}{2}\end{array}$

Thus, the points are $\left(\frac{1}{4},-\frac{1}{2}\right)$ and $\left(\frac{1}{4},\frac{1}{2}\right)$ .

Equation of normal line passing through $\left(\frac{1}{4},\frac{1}{2}\right)$ is $y=-1\left(x-\frac{1}{4}\right)+\frac{1}{2}$ .

The point $\left(a,0\right)$ also lie on the normal line.

  $\begin{array}{c}0=-1\left(a-\frac{1}{4}\right)+\frac{1}{2}\\ 0=-a+\frac{1}{4}+\frac{1}{2}\\ a=\frac{3}{4}\end{array}$

Third coordinate is $\left(\frac{3}{4},0\right)$ .