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

MATH110 December 2012

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Question 08
Find the equations of both lines tangent to the curve $y={\frac {1}{x}}$ which are parallel to the line $y=-100x$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Consider a generic point $P=(a,b)$ on the given curve. What is the slope of the tangent line at the point $x=a$? (Your solution should be a function of $a$.)

Hint 2
Remember that slopes of parallel lines are equal.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\displaystyle {}L(x)=y'(a)(x-a)+y(a).$ We know that y=1/x and so $y(a)={\frac {1}{a}}.$ Next we calculate the derivative: $y'=-{\frac {1}{x^{2}}}$ and plugging in the point x=a we get $y'(a)=-{\frac {1}{a^{2}}}.$ 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

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