Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 26

MATH100 A December 2023

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Question 26
Find the equations of all tangent lines to the curve $$(x^{2}+y^{2})^{2}=2(x^{2}-y^{2})$$ that are parallel to the line ${\textstyle y+1=0}$. All equations must be in the form ${\textstyle y=mx+b}$. A picture of the curve is given below. Plot of a lemniscate.

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
If two lines are parallel, it means they have the same slope.

Hint 2

Break the problem down into steps: 1. Find the slope, ${\textstyle m_{1}}$, of the line ${\textstyle y+1=0}$. Where on the graph would this put the tangent lines? Use this to check your final answer. 2. Determine an expression (possibly implicit) for the slope of the tangent line of ${\textstyle (x^{2}+y^{2})^{2}=2(x^{2}-y^{2})}$. 3. Use what you know from the previous hint to say ${\textstyle y'=m_{1}}$ and solve for the points in the expression from Step 2 which satisfy this.

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. If a tangent line is parallel to the line ${\textstyle y+1=0}$, then it must have the same slope. Rearranging the equation of the line to ${\textstyle y=-1}$ we can see that the slope is ${\textstyle 0}$ (i.e. the line is horizontal), so we need to find the points on the curve that correspond to ${\textstyle y'=0}$. Let’s start by differentiating implicitly: $${\begin{aligned}{\frac {d}{dx}}(x^{2}+y^{2})^{2}&={\frac {d}{dx}}2(x^{2}-y^{2})\\2(x^{2}+y^{2})(2x+2yy')&=2(2x-2yy')\end{aligned}}$$ We know that we are looking for points where ${\textstyle y'=0}$; substituting this in gives $$4x(x^{2}+y^{2})=4x,$$ which has solutions given by pairs $(x,y)$ that satisfy ${\textstyle x^{2}+y^{2}=1}$ or $x=0$. First, replacing $x$ with 0 in the equation that defines the curve yields the equation $y^{4}=-2y^{2}$, which only holds if $y=0$. We see from the picture that the curve does not have a tangent line at the origin, so we can disregard the solution $x=0$, so we move onto the second equation, $x^{2}+y^{2}=1$. We now plug this into the original curve and solve for ${\textstyle y}$. We could solve for ${\textstyle x}$, but then there would be extra steps to determine the tangent line, and since we know the tangent line is horizontal, it’s fully defined by the ${\textstyle y}$-intercept. Plugging this into the original curve gives: $${\begin{aligned}(1)^{2}&=2((1-y^{2})-y^{2})\\1&=2(1-2y^{2})\\4y^{2}&=1\\y&=\pm {\frac {1}{2}}\end{aligned}}$$ So, there are two tangent lines: ${\textstyle y=\pm {\frac {1}{2}}}$.