Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 09 (c)

MATH 180 December 2017

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  •  December 2017  •  

Question 09 (c)
The curve $$\left(x^{2}+y^{2}-2x\right)^{2}=x^{2}+y^{2}$$ is known as a Limaçon of Pascal. (c) Explain why, for every point ${\textstyle (x,y)}$ on the Limaçon with ${\textstyle y\neq 0}$ and tangent line of slope ${\textstyle m}$, there must exist another point ${\textstyle (x,y)}$ on the Limaçon with ${\textstyle y\neq 0}$ and tangent line of slope ${\textstyle -m}$.

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
Look for symmetry in the expression of the slope.

Hint 2
What happens when we replace ${\textstyle y}$ by ${\textstyle -y}$?

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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. By part (b), we have $${\dfrac {dy}{dx}}(x,y)={\dfrac {x-2(x-1)(x^{2}+y^{2}-2x)}{y(2x^{2}+2y^{2}-4x-1)}},$$ where we write ${\textstyle {\dfrac {dy}{dx}}(x,y)}$ in place of ${\textstyle {\dfrac {dy}{dx}}}$ to emphasize that it is the slope at the point ${\textstyle (x,y)}$. Note that if a point $(x,y)$ is on the Limaçon, then so is $(x,-y)$. We can easily show this by plugging $(x,-y)$ into the equation of the curve. By Hint 2, we observe that $${\dfrac {dy}{dx}}(x,-y)={\dfrac {x-2(x-1)(x^{2}+(-y)^{2}-2x)}{-y(2x^{2}+2(-y)^{2}-4x-1)}}=-{\dfrac {x-2(x-1)(x^{2}+y^{2}-2x)}{y(2x^{2}+2y^{2}-4x-1)}}=-{\dfrac {dy}{dx}}(x,y).$$ This means that at the point ${\textstyle (x,-y)}$ the slope has the opposite sign of the slope at ${\textstyle (x,y)}$. In other words, if ${\textstyle {\dfrac {dy}{dx}}(x,y)=m}$, then at the other point ${\textstyle (x,-y)}$, the slope is ${\textstyle {\dfrac {dy}{dx}}(x,-y)=-{\dfrac {dy}{dx}}(x,y)=-m}$. The point ${\textstyle (x,-y)}$ is not the same point as ${\textstyle (x,y)}$ because ${\textstyle y\neq 0}$. Also, at the point ${\textstyle (x,-y)}$, the ${\textstyle y}$-coordinate is ${\textstyle -y\neq 0}$. Answer: ${\textstyle {\color {blue}{\text{If the slope is }}m{\text{ at }}(x,y){\text{ and }}y\neq 0,{\text{ then the slope is }}-m{\text{ at }}(x,-y),{\text{ with }}-y\neq 0}}$.