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:

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

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:

${\begin{aligned}{\frac {\rm {d}}{{\rm {d}}x}}(y)&={\frac {\rm {d}}{{\rm {d}}x}}\left({\frac {x-4}{x+4}}\right)\end{aligned}}$

Using the quotient rule, we get:

${\begin{aligned}u(x)&=x-4&u'(x)=1\\v(x)&=x+4&v'(x)=1\end{aligned}}$

So that means:

${\begin{aligned}{\frac {{\rm {d}}{y}}{{\rm {d}}x}}&={\frac {v(x)u'(x)-u(x)v'(x)}{\left(v(x)\right)^{2}}}\\&={\frac {(x+4)\cdot (1)-(x-4)\cdot (1)}{\left(x+4\right)^{2}}}\\&={\frac {8}{\left(x+4\right)^{2}}}\end{aligned}}$

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

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

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

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

At $x=-8$, we have:

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

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

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

and for $x=-8$

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