Chapter 4.2, Problem 58E

To determine

To find: It is possible to draw the three normal from the point $\left(a,0\right)$ to the parabola $x=y^2$ , then a must be greater than $\frac{1}{2}$ .

And, The values of a for which the other two normal are perpendicular.

Answer to Problem 58E

The value if the other two normal are perpendicular is $a=\frac{3}{4}$ .

Explanation of Solution

Given Information: One of the normal is x- axis.The x- intercept of the normal line is $\left(a,0\right)$ .

Calculation:

The derivative $\frac{dy}{dx}$ of the curve is,

  $\begin{array}{c}x=y^2\\ 1=2y\frac{dy}{dx}\\ \frac{dy}{dx}=\frac{1}{2y}\end{array}$

Any point on the curve is of the form $\left(y^2,y\right)$ . Let $y=b$ , then one such arbitrary point is $\left(b^2,b\right)$ . The slope of tangent o the curve is

  $\begin{array}{c}{\left[m_{\text{tangent}}\right]}_{\left(b^2,b\right)}={\left[\frac{dy}{dx}\right]}_{\left(b^2,b\right)}\\ ={\left[\frac{1}{2y}\right]}_{\left(b^2,b\right)}\\ =\frac{1}{2b}\end{array}$

• Here, $m_{\text{tangent}}$ is the slope of the tangent.

The slope of normal is,

  $\begin{array}{c}{m}_{\text{normal}}=-\frac{1}{m_{\text{tangent}}}\\ =-\frac{1}{\frac{1}{2b}}\\ =-2b\end{array}$

The equation of the line of normal is,

  $\begin{array}{c}y-y_1=m_{\text{normal}}\left(x-x_1\right)\\ y-b=-2b\left(x-b^2\right)\\ y-b=-2bx+2b^3\\ y=-2bx+\left(b+2b^3\right)\end{array}$

The x- intercept of the normal line is $\left(a,0\right)$ . Therefore, substitute $x=a,y=0$ in $y=-2bx+\left(b+2b^3\right)$ ,

  $\begin{array}{c}y=-2bx+\left(b+2b^3\right)\\ 0=-2ba+\left(b+2b^3\right)\\ 2ba=b+2b^3\\ a=\frac{b\left(1+2b^2\right)}{2b}\\ =\frac{\left(1+2b^2\right)}{2}\\ =b^2+\frac{1}{2}\end{array}$

Since $b^2>0,b^2+\frac{1}{2}>\frac{1}{2}$ .

This shows $a>\frac{1}{2}$ .

If the point in the first quadrant has coordinates $\left(b^2,b\right)$ , the coordinates in the fourth quadrant will be, $\left(b^2,-b\right)$ .

The slope of tangent in the first quadrant is $-2y=-2b$ .

The slope of tangent in the fourth quadrant is,

  $\begin{array}{c}-2y=-2\left(-b\right)\\ =2b\end{array}$ .

The two slopes are said to be perpendicular if the product is equal to -1,

  $\begin{array}{c}-2b\times 2b=-1\\ -4b^2=-1\\ b^2=\frac{1}{4}\end{array}$

The corresponding value of a is,

  $\begin{array}{c}a=b^2+\frac{1}{2}\\ =\frac{1}{4}+\frac{1}{2}\\ =\frac{3}{4}\end{array}$

Conclusion:

The value if the other two normal are perpendicular is $a=\frac{3}{4}$ .