Science:Math Exam Resources/Courses/MATH100/December 2018/Question 04

MATH100 December 2018

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Question 04
Find the $x$ and $y$ coordinates of all points on the curve given by the equation: $x^{2}+2y^{2}=8$ for which the tangent line at each of those points passes through the point (0, −6).

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
To find the slope of the tangent line at a point of a curve, use implicit differentiation.

Hint 2
Write the equation of the tangent line at an arbitrary point $(x_{0},y_{0})$ on this curve, using the point-slope formula. Let this line go through (0, -6).

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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. Let $(x_{0},y_{0})$ be a point on the curve. Then this point satisfies $x_{0}^{2}+2y_{0}^{2}=8.~~~~(1)$ To find the slope of the tangent line passing through $(x_{0},y_{0})$ , use implicit differentiation. Differentiate both sides of the equation of the curve with respect to $x$ . By the chain rule, $2x+4yy'=0.$ Let $y_{0}'$ be the slope of the tangent line at point $(x_{0},y_{0})$ . Assume $y_{0}\neq 0$ . Then $y'_{0}=-{\frac {x_{0}}{2y_{0}}}.$ Using the point-slope formula, the tangent line equation to the curve at the point $(x_{0},y_{0})$ can be written as $y-y_{0}=-{\frac {x_{0}}{2y_{0}}}(x-x_{0}).$ Let this tangent line pass through $(0,-6)$ . Putting $x=0$ and $y=-6$ , $-6-y_{0}={\frac {x_{0}^{2}}{2y_{0}}}.$ Together with Equation (1) we can solve for the point. $(x_{0},y_{0})=\left({\frac {8}{3}},-{\frac {2}{3}}\right)$ or $\left(-{\frac {8}{3}},-{\frac {2}{3}}\right).$ Now consider if $y_{0}=0$ , then $x_{0}=\pm 2{\sqrt {2}}.$ The slope of the tangent line DNE. It follows that the tangent line is a vertical line, either $x=2{\sqrt {2}}$ or $x=-2{\sqrt {2}}$ . Neither of them passes through the point $(0,-6).$ Therefore, the points on the curve that satisfy the requirements are $\color {blue}\left({\frac {8}{3}},-{\frac {2}{3}}\right),~\left(-{\frac {8}{3}},-{\frac {2}{3}}\right)$ .