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Math Help - Tangent to an ellipse

  1. #1
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    Tangent to an ellipse

    I am helping a friend with a calculus problem he has, he just started his first calculus course while I have done a couple in the past. However, it seems like I'm rusty

    Here is the problem:

    We have an ellipse:

    9\,{x}^{2}+8\,{\it xy}+25\, \left( y-3 \right) ^{2}=36

    And a point not on the ellipse at (2;2).

    We want to know the two points on the ellipse where there is a tangent also going through that point.

    First, I have implicitly differentiated the equation of the ellipse to end up with:

    {y' = \frac {18\,x+8\,y}{-8\,x-50\,y+150}}

    Then, since I know that:

    {\frac {2-y}{2-x} = \frac {18\,x+8\,y}{-8\,x-50\,y+150}}

    I found:

    {y = -({\frac {36\,x-18\,{x}^{2}+16\,y-8\,yx}{-8\,x-50\,y+150}}-2)}

    Now I need to solve the following system for x and y:

    9\,{x}^{2}+8\,{\it xy}+25\, \left( y-3 \right) ^{2}=36

    {y = -({\frac {36\,x-18\,{x}^{2}+16\,y-8\,yx}{-8\,x-50\,y+150}}-2)}

    But I am unable to. My friend is specifically allowed to use a symbolic calculator to solve the problem, but with Maple I get the following garble:

    1 / / 4 3
    { x = - --------------- \1600 RootOf\1600 _Z - 20000 _Z
    \ 18 (32 xy - 65)

    2 2
    + (93997 + 512 xy) _Z + (-191958 - 5776 xy) _Z + 576 xy + 141453

    \ / 4 3 2
    + 11024 xy/^3 - 14800 RootOf\1600 _Z - 20000 _Z + (93997 + 512 xy) _Z

    2 \ /
    + (-191958 - 5776 xy) _Z + 576 xy + 141453 + 11024 xy/^2 + 45897 RootOf\

    4 3 2
    1600 _Z - 20000 _Z + (93997 + 512 xy) _Z + (-191958 - 5776 xy) _Z

    2 \ / 4 3
    + 576 xy + 141453 + 11024 xy/ + 512 RootOf\1600 _Z - 20000 _Z

    2 2
    + (93997 + 512 xy) _Z + (-191958 - 5776 xy) _Z + 576 xy + 141453

    \ \ / 4 3
    + 11024 xy/ xy - 2888 xy - 45279/, y = RootOf\1600 _Z - 20000 _Z

    2 2
    + (93997 + 512 xy) _Z + (-191958 - 5776 xy) _Z + 576 xy + 141453

    \\
    + 11024 xy/ }
    I have tested the two pairs of solution given with both of the equations in my system and they work. I need help to solve that system here and find the two (x;y) pairs. Thank you a lot in advance.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mathieumg View Post
    I am helping a friend with a calculus problem he has, he just started his first calculus course while I have done a couple in the past. However, it seems like I'm rusty

    Here is the problem:

    We have an ellipse:

    9\,{x}^{2}+8\,{\it xy}+25\, \left( y-3 \right) ^{2}=36

    And a point not on the ellipse at (2;2).
    The equation of a line through (2,2) with slope m is:

    y=mx-2m+2

    That this is a tangent to the ellipse is equivalent to asking that:

    9x^2+8x(mx-2m+2)+25(mx-2m+2-2)^2=36\ \ \ ... \ (1)

    has only one root, or rather that:

    x^2(25m^2+8m+9)+x(-100m^2-66m+16)+(100m^2+100m-11)=0

    has a single root. This occurs when the discriminant is zero:

    (-100m^2-66m+16)^2-4(25m^2+8m+9)(100m^2+100m-11)=0

    or:

    1136 m^2+1340 m-163=0

    Solve this and you will have the two slopes and hence the two tangent equations.

    having forund the slope substitute back into equation (1) to solve for the x coordinate of the point of tangency and substitute that into the equation of the tangent line to find y.



    CB
    Last edited by CaptainBlack; October 15th 2009 at 03:46 AM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    This occurs when the discriminant is zero:

    (-100m^2-66m+16)^-4(25m^2+8m+9))+(100m^2+100m-11)=0
    Shouldn't it be this?


    (-100m^2-66m+16)^2-4(25m^2+8m+9)(100m^2+100m-11)=0
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by mathieumg View Post
    Shouldn't it be this?


    (-100m^2-66m+16)^2-4(25m^2+8m+9)(100m^2+100m-11)=0
    Yes, finger trouble

    CB
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  5. #5
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    Then doesn't it mean we'll end up with m^4 and m^3 values?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by mathieumg View Post
    Then doesn't it mean we'll end up with m^4 and m^3 values?
    No they should cancel out

    CB
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