Chapter 12, Problem 10PS

(a)

To determine

To find: the equation of the tangent line to the parabola at $\left(2,4\right)$ .

Answer to Problem 10PS

The equation of the tangent line at $\left(2,4\right)$ is $y=4x-4$ .

Explanation of Solution

Given information:

The equation of parabola is $y=x^2$ and the point $\left(2,4\right)$ .

Calculation:

The slope of the parabola $y=x^2$ at $\left(2,4\right)$ is computed as follows,

The derivative of the function $y=x^2$ at $x=2$ means that the slope of the tangent of the graph of $y=x^2$ at $\left(2,4\right)$ .

The derivative of the parabola $y=x^2$ is $\frac{dy}{dx}=2x$ .

The slope of the parabola $y=x^2$ at $x=2$ is,

  $\begin{array}{c}m=\frac{dy}{dx}|{}_{x=2}\\ =2\left(2\right)\\ =4\end{array}$

Therefore, the slope of the parabola $y=x^2$ at $x=2$ is 4.

The equation of the tangent line is computed as follows,

Substitute $\left(x_1,y_1\right)=\left(2,4\right)\text{ and }m=4$ in $y-y_1=m\left(x-x_1\right)$ ,

  $\begin{array}{l}y-4=4\left(x-2\right)\\ y-4=4x-8\\ y=4x-8+4\\ y=4x-4\end{array}$

Therefore, the equation of the tangent line at $\left(2,4\right)$ is $y=4x-4$ .

(b)

To determine

To find: the equation of the normal line to the parabola $y=x^2$ at $\left(2,4\right)$ and the line intersect the parabola other than $\left(2,4\right)$ .

Answer to Problem 10PS

The equation of the normal line at $\left(2,4\right)$ is $x+4y=18$ and the line intersects the point $\left(-\frac{9}{4},\frac{81}{16}\right)$ other than $\left(2,4\right)$ .

Explanation of Solution

Calculation:

From part (a), the slope of the tangent line of parabola $y=x^2$ at $x=2$ is 4.

Since the normal line is perpendicular to the tangent line, the product of the two slope is $-1$ .

  $\begin{array}{l}{m}_1{m}_2=-1\\ 4\left(m_2\right)=-1\left[\because \text{slope of tangent is 4}\right]\\ m_2=-\frac{1}{4}\end{array}$

The equation of the normal line is computed as follows,

Substitute $\left(x_1,y_1\right)=\left(2,4\right)\text{ and }m=-\frac{1}{4}$ in $y-y_1=m\left(x-x_1\right)$ ,

  $\begin{array}{l}y-4=-\frac{1}{4}\left(x-2\right)\\ 4\left(y-4\right)=-x+2\\ 4y-16=-x+2\\ x+4y=2+16\\ x+4y=18\end{array}$

Therefore, the equation of the normal line at $\left(2,4\right)$ is $x+4y=18$ .

The normal line intersects the parabola other than $\left(2,4\right)$ is computed as follows.

Substitute $y=x^2$ in $x+4y=18$ and simplify the equation.

  $\begin{array}{l}x+4\left(x^2\right)=18\\ 4x^2+x-18=0\end{array}$

Solution of the equation $4x^2+x-18=0$ ,

  $\begin{array}{l}x=\frac{-1\pm \sqrt{1^2-4\left(4\right)\left(-18\right)}}{2\left(4\right)}\\ x=\frac{-1\pm \sqrt{1^2-4\left(4\right)\left(-18\right)}}{2\left(4\right)}\\ x=\frac{-1\pm \sqrt{289}}{8}\\ x=\frac{-1\pm 17}{8}\\ x=-\frac{18}{8}\text{ or }x=2\\ x=-\frac{9}{4}\end{array}$

Substitute $x=-\frac{9}{4}$ in $x+4y=18$ ,

  $\begin{array}{l}-\frac{9}{4}+4y=18\\ 4y=18+\frac{9}{4}\\ 4y=\frac{81}{4}\\ y=\frac{81}{16}\end{array}$

Therefore, the normal line intersects the parabola other than $\left(2,4\right)$ is $\left(-\frac{9}{4},\frac{81}{16}\right)$ .

(c)

To determine

To find: the equation of the tangent and normal line to the parabola at $\left(0,0\right)$ .

Answer to Problem 10PS

The equation of the tangent line at $\left(0,0\right)$ is $y=0$ and the equation of the normal line is $x=0$ .

Explanation of Solution

Calculation:

The slope of the parabola $y=x^2$ at $\left(0,0\right)$ is computed as follows,

The derivative of the function $y=x^2$ at $x=2$ means that the slope of the tangent of the graph of $y=x^2$ at $\left(0,0\right)$ .

The derivative of the parabola $y=x^2$ is $\frac{dy}{dx}=2x$ .

The slope of the parabola $y=x^2$ at $x=0$ is,

  $\begin{array}{c}m=\frac{dy}{dx}|{}_{x=0}\\ =2\left(0\right)\\ =0\end{array}$

Therefore, the slope of the parabola $y=x^2$ at $x=0$ is 0.

The equation of the tangent line is computed as follows,

Substitute $\left(x_1,y_1\right)=\left(0,0\right)\text{ and }m=0$ in $y-y_1=m\left(x-x_1\right)$ ,

  $\begin{array}{l}y-0=0\left(x-0\right)\\ y=0\end{array}$

Therefore, the equation of the tangent line at $\left(0,0\right)$ is $y=0$ .

The tangent line of the parabola at $\left(0,0\right)$ is x -axis, the perpendicular line of the tangent line passing through $\left(0,0\right)$ is y -axis.

That is, the normal of the parabola at $\left(0,0\right)$ is $x=0$ .