Science:Math Exam Resources/Courses/MATH110/April 2012/Question 01

MATH110 April 2012

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Question 01
The curve below is called a lemniscate; it is given by the equation $\displaystyle (x^{2}+y^{2})^{2}=18(x^{2}-y^{2})$ Find the equation of the tangent line at the point $(2,{\sqrt {2}})$ and sketch the tangent line.

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Hint
In order to find the slope of the tangent line you will need to use implicit differentiation.

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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. To find the equation of the tangent line we need a point and a slope. We already have the point, so it remains to find the slope of the tangent line. We will do so by using implicit differentiation. Implicitly differentiating $(x^{2}+y^{2})^{2}=18(x^{2}-y^{2})$ gives: $2(x^{2}+y^{2})(2x+2y\cdot y')=18(2x-2y\cdot y')$ We could solve for $y'$ and then plug in the point $(2,{\sqrt {2}})$ to find the slope. However, it might be easier to plug in the point first and then solve for $y'$, which is what I will do here: ${\begin{aligned}2((2)^{2}+({\sqrt {2}})^{2})(2(2)+2{\sqrt {2}}\cdot y')&=18(2(2)-2{\sqrt {2}}\cdot y')\\8+4{\sqrt {2}}\cdot y'&=12-6{\sqrt {2}}\cdot y'\\4{\sqrt {2}}\cdot y'+6{\sqrt {2}}\cdot y'&=12-8\\(10{\sqrt {2}})y'&=4\\y'&={\frac {2}{5{\sqrt {2}}}}\end{aligned}}$ Using the point slope formula, with slope ${\frac {2}{5{\sqrt {2}}}}$ and point $(2,{\sqrt {2}})$, we get: $y-{\sqrt {2}}={\frac {2}{5{\sqrt {2}}}}(x-2)$ $y={\frac {2}{5{\sqrt {2}}}}x-{\frac {4}{5{\sqrt {2}}}}+{\sqrt {2}}$ as our tangent line.

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