Science:Math Exam Resources/Courses/MATH110/December 2012/Question 08/Solution 1

Consider a point P=(a,b) on the curve where b=y(a). The tangent line through this point will go through (a,b) and have slope y'(a) where y' is the derivative. Therefore the tangent line L(x) will satisfy

${\displaystyle }L(x)=y^{\prime }(a)(x-a)+y(a).$

We know that y=1/x and so

$y(a)=\frac{1}{a}.$

Next we calculate the derivative:

$y^{\prime }=-\frac{1}{x^2}$

and plugging in the point x=a we get

$y^{\prime }(a)=-\frac{1}{a^2}.$

Up to this point we have enough information to compute the tangent line for any point x=a on the curve. However, we are looking for the two points where the slope is parallel (therefore equal) to -100, the slope of the line given in the question. This gives the equation:

$-\frac{1}{a^2}=-100$

Solving for a, we get

$a=\pm \frac{1}{10}.$

Therefore, we need to find the equation of the tangent line where a = -1/10 and a = 1/10.

• When a = -1/10,

$y(a)=y(-1/10)=\frac{1}{-1/10}=-10$

and the slope is $m=-100$ (because the lines are parallel). Therefore, the equation of the tangent line is

${\displaystyle }L(x)=-100(x-(-1/10))+(-10)$

This is a sufficient answer, but if you simplify to y=mx + b form, you find instead that L(x) = -100x -20.

• When a = 1/10,

$y(a)=y(-1/10)=\frac{1}{1/10}=10,$

with slope $m=-100$. Therefore, the equation of the tangent line is

${\displaystyle }L(x)=-100(x-1/10)+10.$

This is a sufficient answer, but if you simplify to y=mx + b form, you find L(x) = -100x +20