Chapter 3.1, Problem 58E

(a)

To determine

To find: The equation of the tangent lines to the parabola through the points.

Answer to Problem 58E

The tangent line of curve at $\left(-1,0\right)$ is $y=-x-1$.

The tangent line of curve at $\left(5,30\right)$ is $y=11x-25$.

Explanation of Solution

Given:

The parabola is $y=x^2+x$.

The parabola through the point is $\left(2,-3\right)$.

Derivative rules:

(1) Power Rule: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

(2) Sum Rule: $\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]=\frac{d}{dx}\left(f\left(x\right)\right)+\frac{d}{dx}\left(g\left(x\right)\right)$

Formula used:

The equation of tangent line at $\left(x_1,y_1\right)$ is, $y-y_1=m\left(x-x_1\right)$ (1)

Here, m is the slope of the tangent line at $\left(x_1,y_1\right)$ and ${m=\frac{dy}{dx}|}_{x=x_1}$.

Calculation:

The derivative of $y$ is $\frac{dy}{dx}$, which is obtained as follows.

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

Apply the sum rule (2).

$\frac{dy}{dx}=\frac{d}{dx}\left(x^2\right)\text{+}\frac{d}{dx}\left(x\right)$

Apply the power rule (1) and simplify the expression.

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

Therefore, the derivative of the parabola $y=x^2+x$ is $2x+1$.

Given parabola is $y=x^2+x$, that is every point of the parabola is $\left(a,a^2+a\right)$ of the form.

Obtain the equation of the tangent line to the parabola $y=x^2+x$ at the point $\left(a,a^2+a\right)$.

The slope of the tangent line at $\left(a,a^2+a\right)$ is,

$\begin{array}{c}m={\frac{dy}{dx}|}_{x=a}\\ =2\left(a\right)+1\\ =2a+1\end{array}$

Thus, the slope of tangent line at $\left(a,a^2+a\right)$ is $m=2a+1$ (2)

Note that, the slope of the line passing through the point $\left(x_1,x_2\right)$ and $\left(y_1,y_2\right)$ is $m=\frac{y_2-y_1}{x_2-x_1}$.

The tangent line passes through the point $\left(2,-3\right)$ and any point $\left(a,a^2+a\right)$ to the curve is,

$\begin{array}{c}m=\frac{a^2+a-\left(-3\right)}{a-2}\\ =\frac{a^2+a+3}{a-2}\end{array}$

Therefore, the slope $m=\frac{a^2+a+3}{a-2}$ (3)

Compare the equations (2) and (3),

$2a+1=\frac{a^2+a+3}{a-2}$

Cross multiply the equation and simplify the terms,

$\begin{array}{l}\left(a-2\right)\left(2a+1\right)=a^2+a+3\\ 2a^2+\cancel{a}-4a-2=a^2+\cancel{a}+3\\ a^2-4a-5=0\end{array}$

Solve the above equation and obtain $a=-1\text{and }a=5$.

Substitute the values $a=-1\text{and }a=5$ in $\left(a,a^2+a\right)$, the points are $\left(-1,0\right)\text{ and }\left(5,30\right)$.

Since the slope of the tangent line at $\left(a,a^2+a\right)$ is $m=2a+1$, the slope at $\left(-1,0\right)\text{ and }\left(5,30\right)$ are $-1\text{ and 11}$, respectively.

Substitute $\left(-1,0\right)$ for $\left(x_1,y_1\right)$ and $-1$ for m in equation (1),

$\begin{array}{l}y-0=-1\left(x+1\right)\\ y=-x-1\end{array}$

The tangent line to the curve at $\left(-1,0\right)$ is $y=-x-1$.

Substitute $\left(5,30\right)$ for $\left(x_1,y_1\right)$ and 11 for m in equation,

$\begin{array}{l}y-30=11\left(x-5\right)\\ y-30=11x-55\\ y=11x-55+30\\ y=11x-25\end{array}$

The tangent line of curve at $\left(5,30\right)$ is $y=11x-25$.

The graph:

The graph of the curve and tangent line as shown below in Figure 1.

From Figure 1, it is observed that the lines $y=11x-25$ and $y=-x-1$ are tangent to the curve $y=x^2+x$ and these two lines are intersect at the point $\left(2,-3\right)$.

(b)

To determine

To show: There is no line through the point $\left(2,7\right)$ that is tangent to the parabola.

Explanation of Solution

The given parabola is $y=x^2+x$, that is every point of the parabola is $\left(a,a^2+a\right)$ of the form.

Note that, the slope of the line passing through the point $\left(x_1,x_2\right)$ and $\left(y_1,y_2\right)$ is $m=\frac{y_2-y_1}{x_2-x_1}$

The tangent line passing through the point $\left(2,7\right)$ and any point to the parabola $\left(a,a^2+a\right)$,

$\begin{array}{c}m=\frac{a^2+a-\left(7\right)}{a-2}\\ =\frac{a^2+a-7}{a-2}\end{array}$

Therefore, the slope of the tangent line $m=\frac{a^2+a-7}{a-2}$.

From part (a), the slope of tangent line at $\left(a,a^2+a\right)$ is, $m=2a+1$.

Then, the equation becomes,

$\begin{array}{l}2a+1=\frac{a^2+a-7}{a-2}\\ \left(2a+1\right)\left(a-2\right)=a^2+a-7\\ 2a^2-4a+a-2=a^2+a-7\\ a^2-4a+5=0\end{array}$

Note that, the quadratic equation $A{x}^2+Bx+C=0$ has no real solution if the discriminant $B^2-4AC<0$.

$\begin{array}{c}{B}^2-4AC=16-4\left(1\right)\left(5\right)\\ =16-20\\ =-4\end{array}$

Thus, the equation $a^2-4a+5=0$ has no real solution.

Therefore, it is concluded that there is no line such that the tangent line passing through the point $\left(2,7\right)$ to the parabola.

Graph:

The graph of the parabola $y=x^2+x$ and the point $\left(2,7\right)$ is shown below in Figure 2.

From Figure 2, it is observed that there is no tangent line to the parabola through the point $\left(2,7\right)$.