Chapter 3.1, Problem 69E

To determine

To find:

The equation of the tangent and the value of a and b

Answer to Problem 69E

The equation of the value $a=-\frac{1}{2},b=2$

Explanation of Solution

Given:

The tangent to the curve

  $y=a{x}^2$

Parallel to the line

  $2x+y=b$

Concept used:

The equation is in slope −intercept form, $y=mx+c$

An equation for the line through the point $\left(x_1,y_1\right)$ with slope m is

  $y-y_1=m\left(x-x_1\right)$

Calculation:

The function

  $y=a{x}^2\mathrm{.......................}\left(1\right)$

The derivative of a function

  $\begin{array}{l}y=f\left(x\right)\\ y^{\prime }=f^{\prime }\left(x\right)=\frac{dy}{dx}=m\end{array}$

Differentiating the equation (1) with respect to $x$

  $\begin{array}{l}y=a{x}^2\\ y^{\prime }=2ax\mathrm{........................}\left(2\right)\end{array}$

The derivative is slope of the tangent line so in order to the slope of the tangent line

The derivative of constant is zero

  $\begin{array}{l}2x+y=b\\ y=-2x+b\\ y^{\prime }=-\mathrm{2.............................}\left(3\right)\end{array}$

The equation is in slope −intercept form

The slope of the line is $=-2$

From equation (2) and equation (3)

  $\begin{array}{l}-2=2a\left(2\right)\\ -2=4a\\ a=-\frac{1}{2}\end{array}$

The coordinates of points

  $\begin{array}{l}y=a{x}^2\\ y=-\frac{1}{2}\left(2^2\right)\\ y=-2\end{array}$

Solve for b in the line equation

  $\begin{array}{l}2\left(2\right)+\left(-2\right)=b\\ b=2\end{array}$