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Math Help - A line tangent to an implicit function, and through two points not on the function

  1. #1
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    A line tangent to an implicit function, and through two points not on the function

    I have been given the implicit function: X^2+4y^2=36, and have been asked to find a line tangent to it through the point (12,3), which is not on the curve.

    I know there is one horizontal tangent line, at y=3. However, I am having trouble finding the non=horizontal line.

    I know that the derivative is -x/4y. That is fairly simple. I think that I have to set the equation of the line through (12,3) equal to the implicit function, and solve for the slope (which is the derivative) in order to get a numerical value. But I am lost as to how to do that exactly.

    (-x/4y)(x-120+3=x^2+4y^2-36

    Am I on the right track with this?


    Thank you for any help
    Respectfully
    Physics Student
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    We know that the slope of such a line would be

    \frac{3-y}{12-x}=y'(x)=-\frac{x}{4y},

    where (x,y) is a desired point on the curve. Now we may solve for x in terms of y to find the points.
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  3. #3
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    Quote Originally Posted by Scott H View Post
    We know that the slope of such a line would be

    \frac{3-y}{12-x}=y'(x)=-\frac{x}{4y},

    where (x,y) is a desired point on the curve. Now we may solve for x in terms of y to find the points.
    I'm still confused. I can solve for x in terms of y. I know that the equation is going to be

    \frac{-x}{4y}*(x-12)+3

    Using what you gave me, I get y= two rather complicated functions. Should I then plug both of those into the originall implicit and solve for x, then y?

    Thanks for the response
    Physics Student
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    Junior Member Tclack's Avatar
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    My Attempt

    When you rearrange for y, you get y=sqrt[9- (1/4)x^2]

    so as as said before, your slope is

    (3-y)/(12-x) which should equal your derivative -x/4y

    (3-y)/(12-x) = -x/4y

    when you replace your y with sqrt[9- (1/4)x^2] and simplify you should get

    x^2-11x+24 =0

    use quadratic to get x= [11 +/- sqrt(5) ]/2

    That's what I got at least. Although I've been struggling with these problems recently. Does the back of your book give this answer if any?
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    I don't have any solution to the problem that I can check my work against, sorry.
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    Junior Member Tclack's Avatar
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    This is homework I assume. Could you do me a favor and post back with the solution. I'm pretty sure my method was correct, It's always worked for me except for one question (in the thread I JUST started). I really need the mental boost.
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  7. #7
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    Quote Originally Posted by PhysicsStudent View Post
    I have been given the implicit function: X^2+4y^2=36, and have been asked to find a line tangent to it through the point (12,3), which is not on the curve.

    I know there is one horizontal tangent line, at y=3. However, I am having trouble finding the non=horizontal line.

    I know that the derivative is -x/4y. That is fairly simple. I think that I have to set the equation of the line through (12,3) equal to the implicit function, and solve for the slope (which is the derivative) in order to get a numerical value. But I am lost as to how to do that exactly.

    (-x/4y)(x-120+3=x^2+4y^2-36

    Am I on the right track with this?


    Thank you for any help
    Respectfully
    Physics Student
    Let the required tangent be tangent to the ellipse at the point \left(2 \sqrt{9 - a^2}, -a \right) where a > 0 (a simple diagram shows why this is the case).

    Then the gradient of the tangent is m = \frac{\sqrt{9 - a^2}}{2a}.

    But since the tangent passes through the point (12, 3) the gradient is also given by m = \frac{3 + a}{12 - 2 \sqrt{9 - a^2}}.

    Therefore \frac{\sqrt{9 - a^2}}{2a} = \frac{3 + a}{12 - 2 \sqrt{9 - a^2}}.

    Solve this equation for a (remember you want a > 0). I get a = \frac{9}{5}.

    So the required tangent is tangent to the ellipse at the point \left( \frac{24}{5}, \frac{9}{5}\right).

    It should now be simple to get the equation of the required tangent.
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