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Math Help - minimisation question

  1. #1
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    minimisation question

    Suppose that a\geq 0. Find the least distance from the point [tex](a,0) to the ellipse \frac{x^2}{4}+\frac{y^2}{1}=1.

    Well basically i used the distance formula, so D=\sqrt{(x-a)^2+y^2}=\sqrt{(x-a)^2+1-\frac{x^2}{4}}.

    Then i set up a function f(x)=(x-a)^2+1-\frac{x^2}{4}.

    I differentiated it it find f'(x)=2(x-a)-\frac{x}{2}.

    I then found x=\frac{4a}{3} to be the minimum turning point and subbed it back into distance formula to get D=\sqrt{1-\frac{a^2}{3}}.

    However the answer is:

    least distance =\begin{cases}<br />
 &\sqrt{1-\frac{a^2}{3}} \text{ if } 0\leq a\leq 3/2 \\ <br />
 &\left | a-2 \right | \text{ if } a> 3/2<br />
\end{cases}

    i get why its |a-2| if a>2 but i dont get why its |a-2| between a=3/2 and a=2.

    Can someone please explain why?
    Last edited by vuze88; February 5th 2010 at 02:09 AM.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Setting sqrt(1-a^2/3) = 2- a

    we get a = 3/2 this is the point where the point distance from(a,0) to a point on the ellipse is the same as the distance from (a,0) to the vertex(2,0)
    In fact for a = sqrt(3) sqrt(1-a^2/3) = 0 and sqrt(1-a^2/3) is imaginary

    for a > sqrt(3)

    If you want I made an animation demonstrating this on my web site:

    Scratch Paper
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  3. #3
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    hm...i see where you're getting at, but how come the distance formula stops working between a=3/2 and a=sqrt(3). It's still defined between these values. Also, if you let a=1.6 (say), then sqrt(1- (1.6^2 /3)) < |1.6-2| . Shouldnt this be the least distance in this case? And also, how come the distance formula gives 0 when a=sqrt(3). Surely, theres no point on the ellipse that gives this distance.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Note x= 4a/3 at the min pt

    and y= sqrt(1-x^2/4)

    if x = 4a/3 y = sqrt[1 - (4a^2)/9]

    If a = 3/2 x = 2 and y = 0 you are at the vertex

    if a > 3/2 you are no longer on the ellipse and the formula

    D =sqrt(1-a^2/3) is no longer valid as you are not on the ellipse

    you're right as sqrt(3) > 3/2 you don't have to worry about a point on the ellipse whose distance is 0

    thats also why you don't have to worry about a= 1.6 either
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  5. #5
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    thanks for clarifying that
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  6. #6
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    From a=1.5 to 2, the shortest distance to the ellipse is 2-a, or a-2 if a>2.
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