Science:Math Exam Resources/Courses/MATH110/December 2013/Question 02/Solution 1

We first recognise that parallel means that the slopes must be equal. Rearranging the equation, we get:

$\begin{array}{rl}x-2y& =2\\ x-2& =2y\\ y& =\frac{x-2}{2}\\ y& =\frac{1}{2}x-1.\end{array}$

So that means we are looking for points along the curve with slope equal to $\frac{1}{2}$. To get those points, we need the derivative of the curve:

$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}(y)& =\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x-4}{x+4}\right)\end{array}$

Using the quotient rule, we get:

$\begin{array}{rlr}u(x)& =x-4& u^{\prime }(x)=1\\ v(x)& =x+4& v^{\prime }(x)=1\end{array}$

So that means:

$\begin{array}{rl}\frac{\mathrm{d}y}{\mathrm{d}x}& =\frac{v(x)u^{\prime }(x)-u(x)v^{\prime }(x)}{{(v(x))}^2}\\ & =\frac{(x+4)\cdot (1)-(x-4)\cdot (1)}{{(x+4)}^2}\\ & =\frac{8}{{(x+4)}^2}\end{array}$

We are interested in points where the derivative is $\frac{1}{2}$

$\begin{array}{rl}\frac{1}{2}& =\frac{8}{{(x+4)}^2}\\ {(x+4)}^2& =16\\ x+4& =\pm 4\\ x& =0,-8\end{array}$

This means we have to tackle two potential $x$ values. At $x=0$, we have:

$\begin{array}{rl}y& =\frac{0-4}{0+4}=-1\end{array}$

At $x=-8$, we have:

$\begin{array}{rl}y& =\frac{-8-4}{-8+4}=\frac{-12}{-4}=3\end{array}$

So that means the tangent lines are are looking for are: for $x=0$.

$\begin{array}{rl}y& =\frac{1}{2}(x-0)+(-1)\\ & =\frac{1}{2}x-1\\ \end{array}$

and for $x=-8$

$\begin{array}{rl}y& =\frac{1}{2}(x-(-8))+3\\ & =\frac{1}{2}(x+8)+3\\ & =\frac{1}{2}x+7\\ \end{array}$