Science:Math Exam Resources/Courses/MATH102/December 2011/Question 03

MATH102 December 2011

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Question 03
Consider an ellipse given by the equation $\displaystyle {\frac {x^{2}}{9}}+y^{2}=1$ and a point P in the interior with coordinates (4/9, 0). Provide the x and y coordinates of the point(s) on the ellipse that are closest to P. Justify your answer.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Remember that given a point A, you can determine the point B on a curve that is closest to A by determining the point on the curve at which the normal line passes through point A.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. One way to determine the point where the distance between the point P and the ellipse are closest is to determine the equation of the normal line to the ellipse that passes through the point P. We can use implicit differentiation to determine the slope of the tangent line at a point (x,y) on the ellipse: ${\frac {2x}{9}}+2y{\frac {dy}{dx}}=0\quad \rightarrow \quad {\frac {dy}{dx}}=m=-{\frac {x}{9y}}$ Remember that the slope of the normal line to a function at a point x is just the negative reciprocal of the slope at the same point. i.e: if m is slope of the tangent line (as above) and M is the slope of the normal line, both at a point x, then: $M=-{\frac {1}{m}}={\frac {9y}{x}}$ Therefore the equation for the line that is normal to the ellipse at point (a,b) that passes through the point P on the ellipse is given by $(y-0)={\frac {9b}{a}}(x-4/9)$ Since this normal line will also pass through the point where the normal line is being taken, we can solve for (a,b) by solving the following: ${\begin{aligned}(b-0)&={\frac {9b}{a}}(a-4/9)\\a&=9(a-4/9)\\8a&=4\quad \rightarrow \quad a={\frac {1}{2}}\end{aligned}}$ To figure out the corresponding values of b, we use the fact that the point (a,b) must lie on the ellipse: ${\frac {a^{2}}{9}}+b^{2}=1\quad \rightarrow \quad b=\pm {\frac {\sqrt {35}}{6}}$ Notice that we have two solutions! Therefore, there are two points on the ellipse that are closest to the point P. The two points are: ${\color {blue}(a,b)=\left({\frac {1}{2}},{\frac {\sqrt {35}}{6}}\right),\,\left({\frac {1}{2}},-{\frac {\sqrt {35}}{6}}\right)}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This can also be treated as an optimization problem. So, we need to determine the quantity that we want to optimize. We want to determine the point on the given ellipse, $(x,y)$, that is closest to the point $P=(x_{0},y_{0})=(4/9,0)$. Thus, we want to minimize the distance between the two points which we will denote by, $L$. We obtain an expression for $L$ using the formula for distance between two points: ${\begin{aligned}L&={\sqrt {(x-x_{0})^{2}+(y-y_{0})^{2}}}\\&={\sqrt {\left(x-{\frac {4}{9}}\right)^{2}+y^{2}}}\end{aligned}}$ Our constraint in this problem is that the point $(x,y)$ must lie on the ellipse, ${\frac {x^{2}}{9}}+y^{2}=1\quad \rightarrow \quad y^{2}=1-{\frac {x^{2}}{9}}$ Replacing $y^{2}$ in our expression for $L$ with what is given by our constraint, we obtain ${\begin{aligned}L(x)&={\sqrt {\left(x-{\frac {4}{9}}\right)^{2}+1-{\frac {x^{2}}{9}}}}.\end{aligned}}$ To minimize $L$, we need to determine the critical points of $L$. So we compute $L'$ and set it to zero. ${\begin{aligned}L'(x)&={\frac {2(x-4/9)-2x/9}{2{\sqrt {\left(x-4/9\right)^{2}+1-x^{2}/9}}}}={\frac {8x-4}{9{\sqrt {\left(x-4/9\right)^{2}+1-x^{2}/9}}}}=0.\end{aligned}}$ The solution to the above equation is x = 1/2. To figure out the corresponding value(s) of $y$, we substitute the $x$ value into the equation of the ellipse: $y^{2}=1-{\frac {1}{36}}\quad \rightarrow \quad y=\pm {\frac {\sqrt {35}}{6}}.$ So there are two critical points on the ellipse: $(x,y)=\left({\frac {1}{2}},{\frac {\sqrt {35}}{6}}\right),\,\left({\frac {1}{2}},-{\frac {\sqrt {35}}{6}}\right)$ The distance between each of these points and P is the same: $L={\sqrt {\left({\frac {1}{2}}-{\frac {4}{9}}\right)^{2}+\left({\frac {\sqrt {35}}{6}}\right)^{2}}}={\sqrt {\left({\frac {1}{18^{2}}}\right)+\left({\frac {35}{36}}\right)}}\approx 0.988$ We also need to consider the distance between P and the left and right-most points of the ellipse, which occur at (-3,0) and (3,0). Computing value of $L$ for each of these points gives $L$ approximately equal to 3.444 and 2.556, respectively. The distances between P and each of the two endpoints are both larger than the distance between the first two points that we found. Therefore, the (x,y) coordinates of the points that minimize the distance between the ellipse and P are: ${\color {blue}(x,y)=\left({\frac {1}{2}},{\frac {\sqrt {35}}{6}}\right),\,\left({\frac {1}{2}},-{\frac {\sqrt {35}}{6}}\right)}$

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