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Math Help - A question.

  1. #1
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    A question.

    I have this math problem that I am having trouble with, I really don't understand exactly what its asking for, or the best way to do it. I've tried a few different ways, but my answers never seemed right.

    The question is.

    "Find the equation of the line with a positive slope that is the tangent to the ellipse (x^2)/9 + (4^2)/4 = 1 at x=2.

    ??
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  2. #2
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    Quote Originally Posted by Skor View Post
    I have this math problem that I am having trouble with, I really don't understand exactly what its asking for, or the best way to do it. I've tried a few different ways, but my answers never seemed right.

    The question is.

    "Find the equation of the line with a positive slope that is the tangent to the ellipse (x^2)/9 + (y^2)/4 = 1 at x=2.

    ??
    If it's got a positive slope at x = 2, then it's tangent to the bottom side of the ellipse, so let:
    y = - \sqrt{4 \cdot \left ( 1 - \frac{x^2}{9} \right ) } <-- This is the ellipse equation solved for y.

    So
    y(2) = -\frac{2}{3} \sqrt{5}
    meaning the tangent line goes through the point \left ( 2 ,  -\frac{2}{3} \sqrt{5} \right ).

    The tangent line will have a slope equal to the first derivative of y at x = 2:

    m = y' = -\frac{1}{2} \frac{1}{\sqrt{4 \cdot \left ( 1 - \frac{x^2}{9} \right )}} \cdot 4 \frac{-2x}{9} |_{x = 2}

    m = \frac{2x}{9 \sqrt{1 - \frac{x^2}{9}}} |_{x = 2}

    m = \frac{4}{3 \sqrt{5}} = \frac{4 \sqrt{5}}{15}

    So the tangent line is:
    y = \frac{4 \sqrt{5}}{15} x + b

    Using the point \left ( 2 ,  -\frac{2}{3} \sqrt{5} \right ) gives:
    -\frac{2}{3} \sqrt{5} = \frac{8 \sqrt{5}}{15} + b

    b = -\frac{2}{3} \sqrt{5} - \frac{8 \sqrt{5}}{15} = - \frac{18 \sqrt{5}}{15}

    Thus the tangent line is:
    y = \frac{4 \sqrt{5}}{15} x - \frac{18 \sqrt{5}}{15}

    -Dan
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  3. #3
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    Wow thanks for the detailed reply, I think I got this now, thanks alot.
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